Thinking Process

Look at the reading of students $A$ and $B$ given in the question while keeping in mind the concept of precision and accuracy i.e.,

(i) Closeness of reading is precision, and

(ii) If mean of reading is exactly same as the correct value then it is known as accuracy.

Answer

(b) Average of readings of student, $A=\frac{3.01+2.99}{2}=3.00$

Average of readings of student, $B=\frac{3.05+2.95}{2}=3.00$

Correct reading $=3.00$

For both the students, average value is close to the correct value. Hence, readings of both are accurate.

Readings of student $A$ are close to each other (differ only by 0.02 ) and also close to the correct reading, hence, readings of $A$ are precise also. But readings of $B$ are not close to each other (differ by 0.1 ) and hence are not precise.

• (a) Results of both the students are neither accurate nor precise:

  • This is incorrect because the average readings of both students are close to the correct value of 3.0 g, indicating that their results are accurate. However, only student A’s readings are precise, as they are close to each other.

• (c) Results of student $B$ are neither precise nor accurate:

  • This is incorrect because the average reading of student B is close to the correct value of 3.0 g, indicating that the results are accurate. However, the readings are not precise as they differ by 0.1 g.

• (d) Results of student $B$ are both precise and accurate:

  • This is incorrect because, although the average reading of student B is accurate, the individual readings differ by 0.1 g, indicating that the results are not precise.

Answer

(c) There are three common scales to measure temperature ${ }^{\circ} \mathrm{C}$ (degree celsius),

${ }^{\circ} \mathrm{F}$ (degree fahrenheit) and $\mathrm{K}$ (kelvin). The $\mathrm{K}$ is the $\mathrm{SI}$ unit.

The temperatures on two scales are related to each other by the following relationship.

$$ { }^{\circ} \mathrm{F}=\frac{9}{5} t^{\circ} \mathrm{C}+32 $$

Putting the values in above equation

$$ \begin{array}{rlrl} & 200-32 =\frac{9}{5} t^{\circ} \mathrm{C} \\ \Rightarrow & \frac{9}{5} t^{\circ} \mathrm{C} =168 \\ \Rightarrow & t^{\circ} \mathrm{C} =\frac{168 \times 5}{9}=93.3^{\circ} \mathrm{C} \end{array} $$

• Option (a) $40{ }^{\circ} \mathrm{C}$: This is incorrect because when converting $200^{\circ} \mathrm{F}$ to Celsius using the formula $( t^{\circ} \mathrm{C} = \frac{5}{9} (200 - 32))$, the result is not $40{ }^{\circ} \mathrm{C}$ but $93.3{ }^{\circ} \mathrm{C}$.

• Option (b) $94{ }^{\circ} \mathrm{C}$: This is incorrect because the precise conversion of $200^{\circ} \mathrm{F}$ to Celsius is $93.3{ }^{\circ} \mathrm{C}$, not $94{ }^{\circ} \mathrm{C}$. The slight difference is due to rounding errors.

• Option (d) $30^{\circ} \mathrm{C}$: This is incorrect because $30^{\circ} \mathrm{C}$ is far too low for the conversion of $200^{\circ} \mathrm{F}$. The correct conversion is $93.3{ }^{\circ} \mathrm{C}$.

Answer

(c) Since, molarity $(M)$ is calculated by following equation

$$ \begin{aligned} \text { Molarity } & =\frac{\text { weight } \times 1000}{\text { molecular weight } \times \text { volume }(\mathrm{mL})} \\ & =\frac{5.85 \times 1000}{58.5 \times 500}=0.2 \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned} $$

Note Molarity of solution depends upon temperature because volume of a solution is temperature dependent.

• Option (a) $4 \mathrm{~mol} \mathrm{~L}^{-1}$: This option is incorrect because the calculated molarity of the solution is $0.2 \mathrm{~mol} \mathrm{~L}^{-1}$, not $4 \mathrm{~mol} \mathrm{~L}^{-1}$. The value $4 \mathrm{~mol} \mathrm{~L}^{-1}$ is significantly higher than the correct molarity.

• Option (b) $20 \mathrm{~mol} \mathrm{~L}^{-1}$: This option is incorrect because the calculated molarity of the solution is $0.2 \mathrm{~mol} \mathrm{~L}^{-1}$, not $20 \mathrm{~mol} \mathrm{~L}^{-1}$. The value $20 \mathrm{~mol} \mathrm{~L}^{-1}$ is extremely high and unrealistic for the given amount of solute and volume.

• Option (d) $2 \mathrm{~mol} \mathrm{~L}^{-1}$: This option is incorrect because the calculated molarity of the solution is $0.2 \mathrm{~mol} \mathrm{~L}^{-1}$, not $2 \mathrm{~mol} \mathrm{~L}^{-1}$. The value $2 \mathrm{~mol} \mathrm{~L}^{-1}$ is ten times higher than the correct molarity.

Thinking Process

In case of solution, molarity is calculated by using molarity equation, $M_{1} V_{1}=M_{2} V_{2}$, we have, $V_{1}$ (before dilution) and $V_{2}$ (after dilution), so calculate molarity of the given solution from this equation.

Answer

(b) Given that,

$$ \begin{aligned} M_{1} & =5 \mathrm{M} \\ V_{1} & =500 \mathrm{~mL} \\ V_{2} & =1500 \mathrm{~mL} \\ M_{2} & =M \end{aligned} $$

For dilution, a general formula is

$$ \begin{aligned} M_{1} V_{1} & =M_{2} V_{2} \\ \text { (Before dilution) } & \text { (After dilution) } \\ 500 \times 5 \mathrm{M} & =1500 \times M \\ M= & \frac{5}{3}=1.66 \mathrm{M} \end{aligned} $$

• Option (a) $1.5 \mathrm{M}$: This option is incorrect because it does not match the calculated molarity after dilution. The correct calculation using the dilution formula $( M_1 V_1 = M_2 V_2 ) gives ( M_2 = 1.66 \mathrm{M} ), not ( 1.5 \mathrm{M})$.

• Option (c) $0.017 \mathrm{M}$: This option is incorrect because it is significantly lower than the calculated molarity. The correct molarity after dilution is $( 1.66 \mathrm{M} ), and ( 0.017 \mathrm{M} )$ is not a plausible result given the initial conditions.

• Option (d) $1.59 \mathrm{M}$: This option is incorrect because it is close but not accurate. The precise calculation using the dilution formula results in $( 1.66 \mathrm{M} ), not ( 1.59 \mathrm{M} )$.

Thinking Process

The number of atoms is related to Avogadro’s number $\left(N_{A}\right)$ by

Number of atoms $=$ moles $\times N_{A}$

The number of atoms of elements can be compared easily on the basis of their moles only because $N_{A}$ is a constant value. Thus, element with large number of moles will possess greatest number of atoms.

Answer

(d) For comparing number of atoms, first we calculate the moles as all are monoatomic and hence, moles $\times N_{A}=$ number of atoms.

$$ \begin{aligned} \text { Moles of } 4 \mathrm{~g} \mathrm{He} & =\frac{4}{4}=1 \mathrm{~mol} \\ 46 \mathrm{~g} \mathrm{Na} & =\frac{46}{23}=2 \mathrm{~mol} \\ 0.40 \mathrm{~g} \mathrm{Ca} & =\frac{0.40}{40}=0.1 \mathrm{~mol} \\ 12 \mathrm{~g} \mathrm{He} & =\frac{12}{4}=3 \mathrm{~mol} \end{aligned} $$

Hence, $12 \mathrm{~g}$ He contains greatest number of atoms as it possesses maximum number of moles.

• (a) $4 \mathrm{~g}$ He: This option is incorrect because $4 \mathrm{~g}$ of He corresponds to $1 \mathrm{~mol}$, which is fewer moles compared to $12 \mathrm{~g}$ of He, which corresponds to $3 \mathrm{~mol}$.

• (b) $46 \mathrm{~g} \mathrm{Na}$: This option is incorrect because $46 \mathrm{~g}$ of Na corresponds to $2 \mathrm{~mol}$, which is fewer moles compared to $12 \mathrm{~g}$ of He, which corresponds to $3 \mathrm{~mol}$.

• (c) $0.40 \mathrm{~g} \mathrm{Ca}$: This option is incorrect because $0.40 \mathrm{~g}$ of Ca corresponds to $0.01 \mathrm{~mol}$, which is significantly fewer moles compared to $12 \mathrm{~g}$ of He, which corresponds to $3 \mathrm{~mol}$.

Answer

(c) In the given question, $0.9 \mathrm{~g} \mathrm{~L}^{-1}$ means that $1000 \mathrm{~mL}$ (or $1 \mathrm{~L}$ ) solution contains $0.9 \mathrm{~g}$ of glucose.

$\therefore$ Number of moles $=0.9 \mathrm{~g}$ glucose $=\frac{0.9}{180}$ mol glucose

$$ =5 \times 10^{-3} \mathrm{~mol} \text { glucose } $$

(where, molecular mass of glucose $\left(C_6 H_12 \mathrm{O}_{6}\right)=12 \times 6+12 \times 1+6 \times 16=180 \mathrm{u}$ ) i.e., $1 \mathrm{~L}$ solution contains 0.05 mole glucose or the molarity of glucose is $0.005 \mathrm{M}$.

• Option (a) $5 \mathrm{M}$: This option is incorrect because it suggests a much higher concentration of glucose than calculated. Given that $0.9 \mathrm{~g}$ of glucose in $1 \mathrm{~L}$ of solution corresponds to $0.005 \mathrm{M}$, a concentration of $5 \mathrm{M}$ would imply $900 \mathrm{~g}$ of glucose per liter, which is not the case.

• Option (b) $50 \mathrm{M}$: This option is incorrect because it suggests an even higher concentration of glucose than option (a). A concentration of $50 \mathrm{M}$ would imply $9000 \mathrm{~g}$ of glucose per liter, which is far beyond the given $0.9 \mathrm{~g}$ per liter.

• Option (d) $0.5 \mathrm{M}$: This option is incorrect because it also suggests a higher concentration than calculated. A concentration of $0.5 \mathrm{M}$ would imply $90 \mathrm{~g}$ of glucose per liter, which is significantly more than the given $0.9 \mathrm{~g}$ per liter.

Answer

(d) Molality is defined as the number of moles of solute present in $1 \mathrm{~kg}$ of solvent. It is denoted by $m$.

Thus,

$$ \text { Molality }(m)=\frac{\text { Moles of solute }}{\text { Mass of solvent (in } \mathrm{kg})} \quad…(i) $$

Given that, Mass of solvent $\left(\mathrm{H}_{2} \mathrm{O}\right)=500 \mathrm{~g}=0.5 \mathrm{~kg}$

$$ \text { Weight of } \mathrm{HCl}=18.25 \mathrm{~g} $$

Molecular weight of $\mathrm{HCl}=1 \times 1+1 \times 35.5=36.5 \mathrm{~g}$

$$ \begin{aligned} \text { Moles of } \mathrm{HCl} & =\frac{18.25}{36.5}=0.5 \\ m & =\frac{0.5}{0.5}=1 \mathrm{~m} \quad[\text{from Eq. (i)}] \end{aligned} $$

• Option (a) $0.1 \mathrm{~m}$: This option is incorrect because the calculation of molality is based on the number of moles of solute per kilogram of solvent. Given that the moles of HCl are 0.5 and the mass of the solvent is 0.5 kg, the molality is calculated as $( \frac{0.5}{0.5} = 1 \mathrm{~m} )$, not 0.1 m.

• Option (b) $1 \mathrm{M}$: This option is incorrect because it refers to molarity, not molality. Molarity is the number of moles of solute per liter of solution, whereas molality is the number of moles of solute per kilogram of solvent. The problem specifically asks for molality, not molarity.

• Option (c) $0.5 \mathrm{~m}$: This option is incorrect because the calculation of molality shows that the number of moles of HCl is 0.5 and the mass of the solvent is 0.5 kg. Therefore, the molality is $( \frac{0.5}{0.5} = 1 \mathrm{~m})$, not 0.5 m.

Answer

(a) One mole of any substance contains $6.022 \times 10^{23}$ atoms $/$ molecules.

Hence, Number of millimoles of $ H_2 SO_4 $

$$ \begin{aligned} & =\text { molarity } \times \text { volume in } \mathrm{mL} \\ & =0.02 \times 100=2 \text { millimoles } \\ & =2 \times 10^{-3} \mathrm{~mol} \end{aligned} $$

Number of molecules $=$ number of moles $\times N_{\mathrm{A}}$

$$ \begin{aligned} & =2 \times 10^{-3} \times 6.022 \times 10^{23} \\ & =12.044 \times 10^{20} \text { molecules } \end{aligned} $$

• Option (b) $6.022 \times 10^{23}$ molecules is incorrect because this value represents the number of molecules in one mole of a substance, not in 100 mL of a 0.02 M solution.

• Option (c) $1 \times 10^{23}$ molecules is incorrect because it underestimates the number of molecules present in 100 mL of a 0.02 M solution. The correct calculation yields $12.044 \times 10^{20}$ molecules.

• Option (d) $12.044 \times 10^{23}$ molecules is incorrect because it overestimates the number of molecules present in 100 mL of a 0.02 M solution. The correct calculation yields $12.044 \times 10^{20}$ molecules.

Answer

(b) Molecular mass of $\mathrm{CO}_{2}=1 \times 12+2 \times 16=44 \mathrm{~g}$

$1 \mathrm{~g}$ molecule of $\mathrm{CO}_{2}$ contains $1 \mathrm{~g}$ atoms of carbon

Q $\quad 44 \mathrm{~g}$ of $\mathrm{CO}_{2}$ contain $\mathrm{C}=12 \mathrm{~g}$ atoms of carbon

$\therefore \quad$% of $\mathrm{C}$ in $\mathrm{CO}_{2}=\frac{12}{44} \times 100=27.27 $%

Hence, the mass per cent of carbon in $\mathrm{CO}_{2}$ is $27.27 $%.

• Option (a) $0.034$ %: This value is far too low to represent the mass percent of carbon in carbon dioxide. Given that carbon dioxide is composed of one carbon atom and two oxygen atoms, the mass percent of carbon must be significantly higher than this value.

• Option (c) $3.4$ %: This value is also too low. The mass percent of carbon in carbon dioxide is calculated based on the ratio of the mass of carbon to the total mass of the molecule. Since the molecular mass of CO₂ is 44 g/mol and the mass of carbon is 12 g/mol, the correct mass percent is much higher than 3.4%.

• Option (d) $28.7$ %: This value is slightly higher than the correct mass percent. The correct calculation shows that the mass percent of carbon in carbon dioxide is 27.27%, not 28.7%. This discrepancy indicates an overestimation.

Thinking Process

(i) Empirical formula shows that number of moles of different elements present in a molecule, so find the number of moles by dividing molecular mass with empirical formula mass.

(ii) To calculate the molecular formula of the compound, multiply the number of moles with empirical formula.

Answer

(c) Empirical formula mass $=\mathrm{CH}_{2} \mathrm{O}$

$$ \begin{aligned} & =12+2 \times 1+16=30 \\ \text { Molecular mass } & =180 \\ n & =\frac{\text { Molecular mass }}{\text { Empirical formula mass }} \\ & =\frac{180}{30}=6 \end{aligned} $$

$\therefore \quad$ Molecular formula $=n \times$ empirical formula

$$ =6 \times CH_2 O \\ =C_6 H_12 O_6 $$

• Option (a) $C_9 H_{18} O_9$: This option is incorrect because the molecular formula derived from the empirical formula $\mathrm{CH} _{2} \mathrm{O}$ and the given molecular mass of $(180 \mathrm{~g})$ results in $(C _6 H _{12} O_6$, not $C_9 H _{18} O _9)$. The factor $n$ calculated is 6, not 9.

• Option (b) $CH_2 O$: This option is incorrect because $CH_2 O$ is the empirical formula, not the molecular formula. The molecular formula must be a multiple of the empirical formula that matches the given molecular mass of $(180 \mathrm{~g})$. The correct multiple is 6, resulting in $C _6 H _{12} O_6$.

• Option (d) $C _{2} H _{4} O _{2}$: This option is incorrect because it does not match the molecular mass of $(180 \mathrm{~g})$. The molecular formula derived from the empirical formula $(\mathrm{CH} _{2} \mathrm{O})$ and the given molecular mass results in $(C _6 H _{12} O_6$, not $C _{2} H _{4} O _{2})$. The factor $n$ calculated is 6, not 2.

Answer

(a) Given that, density of solution $=3.12 \mathrm{~g} \mathrm{~mL}^{-1}$

Volume of solution $=1.5 \mathrm{~mL}$

For a solution,

$$ \begin{aligned} \text { Mass } & =\text { volume } \times \text { density } \\ & =1.5 \mathrm{~mL} \times 3.12 \mathrm{~g} \mathrm{~mL}^{-1}=4.68 \mathrm{~g} \end{aligned} $$

The digit 1.5 has only two significant figures, so the answer must also be limited to two significant figures. So, it is rounded off to reduce the number of significant figures.

Hence, the answer is reported as $4.7 \mathrm{~g}$.

• Option (b) $4680 \times 10^{-3} \mathrm{~g}$: This option is incorrect because it is not expressed in the correct significant figures. The correct answer should be rounded to two significant figures, and $4680 \times 10^{-3} \mathrm{~g}$ (which equals $4.680 \mathrm{~g}$) has four significant figures.

• Option (c) $4.680 \mathrm{~g}$: This option is incorrect because it has four significant figures. The correct answer should be rounded to two significant figures, and $4.680 \mathrm{~g}$ is not rounded to the appropriate number of significant figures.

• Option (d) $46.80 \mathrm{~g}$: This option is incorrect because it is not only expressed with four significant figures but also represents an incorrect value. The correct calculation yields $4.68 \mathrm{~g}$, not $46.80 \mathrm{~g}$.

Answer

(c) A compound is a pure substance containing two or more than two elements combined together in a fixed proportion by mass and which can be decomposed into its constituent elements by suitable chemical methods.

Further, the properties of a compound are quite different from the properties of constituent elements. e.g., water is a compound containing hydrogen and oxygen combined together in a fixed proportionation. But the properties of water are completely different from its constituents, hydrogen and oxygen.

• (a) A molecule of a compound has atoms of different elements: This statement is correct. A compound is formed when atoms of different elements chemically combine in a fixed ratio.

• (b) A compound cannot be separated into its constituent elements by physical methods of separation: This statement is correct. Compounds can only be separated into their constituent elements by chemical methods, not by physical methods.

• (d) The ratio of atoms of different elements in a compound is fixed: This statement is correct. In a compound, the elements are always present in a definite proportion by mass.

Thinking Process

This problem is based upon the law of conservation of mass as well as limiting reagent.

(i) Law of conservation of mass is that in which total mass of reactants is equal to total mass of products.

(ii) Limiting reagent represents the reactant which reacts completely in the reaction.

Answer

(a) According to the law of conservation of mass,

Total mass of reactants $=$ Total mass of products

Amount of $Fe _{2} O _{3}$ is decided by limiting reagent.

• (b) The law of multiple proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers. This reaction does not illustrate the law of multiple proportions; it illustrates the law of conservation of mass.

• (c) The amount of $( Fe_{2}O_{3} )$ produced is determined by the limiting reagent, not by taking any one of the reactants in excess. Taking one reactant in excess will not increase the amount of $( Fe_{2}O_{3} )$ produced beyond what the limiting reagent allows.

• (d) Taking any one of the reactants (iron or oxygen) in excess will not decrease the amount of $( Fe_{2}O_{3} )$ produced. The amount of $( Fe_{2}O_{3} )$ produced is determined by the limiting reagent, and taking one reactant in excess will not affect the amount produced by the limiting reagent.

Answer

(b) In this equation,

$$ \underset{44 ~g}{C_3 H_8(g)}+\underset{32 ~g}{O_2(g)} \longrightarrow \underset{44 ~g}{CO_2(g)}+{\underset{18 ~g}{H_2} O}(g) $$

i.e., mass of reactants $\neq$ mass of products.

Hence, law of conservation of mass is not followed.

• For option (a): The reaction is $2 Mg(s) + O_2(g) \longrightarrow 2 MgO(s)$. The molar masses are:

  • $2 \times 24.305 , \text{g/mol} = 48.61 , \text{g}$ for $2 , \text{Mg}$
  • $1 \times 32.00 , \text{g/mol} = 32.00 , \text{g}$ for $O_2$
  • $2 \times 40.305 , \text{g/mol} = 80.61 , \text{g}$ for $2 , \text{MgO}$ The total mass of reactants is $48.61 , \text{g} + 32.00 , \text{g} = 80.61 , \text{g}$, which equals the total mass of products. Hence, the law of conservation of mass is followed.

• For option (c): The reaction is $P_4(s) + 5 O_2(g) \longrightarrow P_4O_{10}(s)$. The molar masses are:

  • $1 \times 123.895 , \text{g/mol} = 123.895 , \text{g}$ for $P_4$
  • $5 \times 32.00 , \text{g/mol} = 160.00 , \text{g}$ for $5 , O_2$
  • $1 \times 283.895 , \text{g/mol} = 283.895 , \text{g}$ for $P_4O_{10}$ The total mass of reactants is $123.895 , \text{g} + 160.00 , \text{g} = 283.895 , \text{g}$, which equals the total mass of products. Hence, the law of conservation of mass is followed.

• For option (d): The reaction is $CH_4(g) + 2 O_2(g) \longrightarrow CO_2(g) + 2 H_2O(g)$. The molar masses are:

  • $1 \times 16.04 , \text{g/mol} = 16.04 , \text{g}$ for $CH_4$
  • $2 \times 32.00 , \text{g/mol} = 64.00 , \text{g}$ for $2 , O_2$
  • $1 \times 44.01 , \text{g/mol} = 44.01 , \text{g}$ for $CO_2$
  • $2 \times 18.02 , \text{g/mol} = 36.04 , \text{g}$ for $2 , H_2O$ The total mass of reactants is $16.04 , \text{g} + 64.00 , \text{g} = 80.04 , \text{g}$, which equals the total mass of products. Hence, the law of conservation of mass is followed.

Answer

(b) The element, carbon, combines with oxygen to form two compounds, namely, carbon dioxide and carbon monoxide. In $\mathrm{CO}_{2}, 12$ parts by mass of carbon combine with 32 parts by mass of oxygen while in CO, 12 parts by mass of carbon combine with 16 parts by mass of oxygen.

Therefore, the masses of oxygen combine with a fixed mass of carbon (12 parts) in $\mathrm{CO}_{2}$ and $\mathrm{CO}$ are 32 and 16 respectively. These masses of oxygen bear a simple ratio of $32: 16$ or $2: 1$ to each other.

This is an example of law of multiple proportion

• (a) This statement describes the law of definite proportions, not the law of multiple proportions. The law of definite proportions states that a chemical compound always contains its component elements in a fixed ratio by mass, regardless of the source of the compound.

• (c) This statement describes the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. The amount of magnesium taken for the reaction being equal to the amount of magnesium in magnesium oxide formed is an example of this law.

• (d) This statement describes Gay-Lussac’s law of combining volumes, which states that when gases react together at constant temperature and pressure, the volumes of the reacting gases and the volumes of the products (if gaseous) are in simple whole number ratios.

Answer

$(a, d)$

1 mole of $ O_{2}$ gas at STP $=6.022 \times 10^{23}$ molecules of $ O_{2}$ (Avogadro number) $=32 ~g {\text { of } {O}_{2}}$

Hence, 1 mole of oxygen gas is equal to molecular weight of oxygen as well as Avogadro number.

• Option (b) is incorrect because 1 mole of oxygen gas (O₂) consists of molecules, not individual atoms. Therefore, 1 mole of O₂ contains $6.022 \times 10^{23}$ molecules, which corresponds to $2 \times 6.022 \times 10^{23}$ atoms of oxygen, not $6.022 \times 10^{23}$ atoms.

• Option (c) is incorrect because the molar mass of oxygen gas (O₂) is 32 grams per mole, not 16 grams. 16 grams would correspond to half a mole of O₂, not a full mole.

Answer

( $b, c$ )

For the reaction, $\quad H_{2} SO_{4}+2 NaOH \longrightarrow Na_{2} SO_{4}+2 H_{2} O$

$1 ~L $ of $0.1 M H_{2} SO_{4}$ contains $=0.1$ mole of $H_{2} SO_{4}$

$1 ~L$ of $0.1 M NaOH $ contains $=0.1$ mole of $NaOH $

According to the reaction, 1 mole of $ H_2 SO_4 $ reacts with 2 moles of $NaOH $. Hence, 0.1 mole of $NaOH$ will react with 0.05 mole of $H_{2} SO_{4}$ (and 0.05 mole of $H_{2} SO_{4}$ will be left unreacted), i.e., $NaOH$ is the limiting reactant. Since, 2 moles of $NaOH$ produce 1 mole of $Na_{2} SO_{4}$.

Hence, 0.1 mole of $NaOH$ will produce 0.05 mole of $Na_{2} SO_{4}$.

$\begin{aligned} \text{Mass of} \quad Na_{2} SO_{4} & =\text { moles } \times \text { molar mass } \\ & =0.5 \times(46+32+64) g \\ & =7.10 ~g \\ \text { Volumme of solution after mixing } & =2 ~L\end{aligned}$

Since, only 0.05 mole of $H_{2} SO_{4}$ is left behind as $NaOH$ completely used in the reaction. Therefore, molarity of the given solution is calculated from moles of $H_{2} SO_{4}$.

$H_{2} SO_{4}$ left unreacted in the solution $=0.05$ mole

$\therefore \quad$ Molarity of the solution $=\frac{0.05}{2}=0.025 ~mol ~L^{-1}$

• Option (a) $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$: This option is incorrect because it suggests that the molarity of the sodium sulphate formed is $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$. However, the correct molarity of sodium sulphate in the final solution is $0.025 \mathrm{~mol} \mathrm{~L}^{-1}$, as calculated from the reaction stoichiometry and the total volume of the solution.

• Option (d) $3.55 \mathrm{~g}$: This option is incorrect because it suggests that the mass of sodium sulphate formed is $3.55 \mathrm{~g}$. However, the correct mass of sodium sulphate formed is $7.10 \mathrm{~g}$, as calculated from the number of moles of sodium sulphate produced and its molar mass.

Answer

$(c, d)$

(c) Number of atoms in $28 \mathrm{~g}$ of $N_{2}=\frac{28}{28} \times N_{A} \times 2=2 N_{A}$ (where, $N_{A}=$ Avogadro number)

Number of atoms in $32 g$ of $O_2 = \frac{32}{32} \times N_A \times 2 = 2N_A $

(d) $12 \mathrm{~g}$ of $\mathrm{C}(\mathrm{s})$ contains atoms $=\frac{12}{12} \times N_{\mathrm{A}} \times 1=\mathrm{N}_{\mathrm{A}}$

Number of atoms in $23 \mathrm{~g}$ of $\mathrm{Na}(s)=\frac{23}{23} \times N_{A} \times 1=N_{A}$

• (a) $16 ~g$ of $O_{2}(~g)$ and $4 ~g$ of $H_{2}(~g)$:

  • Number of atoms in $16 ~g$ of $O_{2} = \frac{16}{32} \times N_A \times 2 = N_A$
  • Number of atoms in $4 ~g$ of $H_{2} = \frac{4}{2} \times N_A \times 2 = 4N_A$
  • The number of atoms in $16 ~g$ of $O_{2}$ is $N_A$, while the number of atoms in $4 ~g$ of $H_{2}$ is $4N_A$. Therefore, they do not have the same number of atoms.

• (b) $16 ~g$ of $O_{2}$ and $44 ~g$ of $CO_{2}$:

  • Number of atoms in $16 ~g$ of $O_{2} = \frac{16}{32} \times N_A \times 2 = N_A$
  • Number of atoms in $44 ~g$ of $CO_{2} = \frac{44}{44} \times N_A \times 3 = 3N_A$
  • The number of atoms in $16 ~g$ of $O_{2}$ is $N_A$, while the number of atoms in $44 ~g$ of $CO_{2}$ is $3N_A$. Therefore, they do not have the same number of atoms.

Answer

$(a, b)$

(a) Molarity $(M)=\frac{\text { weight of } \mathrm{NaOH} \times 1000}{\text { Molecular weight of } \mathrm{NaOH} \times V(\mathrm{~mL})}$

$$ =\frac{20 \times 1000}{40 \times 200}=2.5 \mathrm{M} $$

(b) $M=\frac{0.5 \times 1000}{200}=2.5 \mathrm{M}$

(c) $M=\frac{40 \times 1000}{10 \times 100}=10 \mathrm{M}$

(d) $M=\frac{20 \times 1000}{56 \times 200}=1.785 \mathrm{M}$

Thus, $20 \mathrm{~g} \mathrm{NaOH}$ in $200 \mathrm{~mL}$ of solution and $0.5 \mathrm{~mol}$ of $\mathrm{KCl}$ in $200 \mathrm{~mL}$ have the same concentration.

• Option (c): The concentration is calculated as 10 M, which is significantly higher than the concentration of 2.5 M found in options (a) and (b).

• Option (d): The concentration is calculated as 1.785 M, which is lower than the concentration of 2.5 M found in options (a) and (b).

Answer

(c, d)

The number of molecules can be calculated as follows

Number of molecules $=\frac{\text { Mass }}{\text { Molar mass }} \times$ Avogadro number $\left(N_{A}\right)$

Number of molecules, in $16 ~g$ oxygen $=\frac{16}{32} \times N_{A}=\frac{N_{A}}{2}$

In $16 ~g$ of $CO=\frac{16}{28} \times N_{A}=\frac{N_{A}}{1.75}$

In $28 ~g$ of $N_{2}=\frac{28}{28} \times N_{A}=N_{A}$

In $14 ~g$ of $N_{2}=\frac{14}{28} \times N_{A}=\frac{N_{A}}{2}$

In $1 ~g$ of $H_{2}=\frac{1}{2} \times N_{A}=\frac{N_{A}}{2}$

So, $16 ~g$ of $O_{2}=14 ~g$ of $N_{2}=1.0 ~g$ of $H_{2}$

• For option (a) $16 \mathrm{~g}$ of $\mathrm{CO}$: The number of molecules in $16 \mathrm{~g}$ of $\mathrm{CO}$ is calculated as $\frac{16}{28} \times N_{A} = \frac{N _{A}}{1.75}$. This is not equal to the number of molecules in $16 \mathrm{~g}$ of $\mathrm{O} _{2}$, which is $\frac{N _{A}}{2}$.

• For option (b) $28 \mathrm{~g}$ of $\mathrm{N} _{2}$: The number of molecules in $28 \mathrm{~g}$ of $\mathrm{N} _{2}$ is calculated as $\frac{28}{28} \times N _{A} = N _{A}$. This is not equal to the number of molecules in $16 \mathrm{~g}$ of $(\mathrm{O} _{2}$, which is $\frac{N _{A}}{2})$.

Answer

$(c, d)$

Both mole fraction and mass per cent are unitless as both are ratios of moles and mass respectively.

$$ \begin{aligned} \text { Mole fraction } & =\frac{\text { Number of moles of solute }}{\text { Number of moles of solution }}=\frac{\text { moles }}{\text { moles }} \\ & =\frac{\text { Number of moles of solvent }}{\text { Number of moles of solution }}=\frac{\text { moles }}{\text { moles }} \\ \text { Mass per cent } & =\frac{\text { Mass of solute in gram }}{\text { Mass of solution in gram }} \times 100 \end{aligned} $$

• Molality: Molality is defined as the number of moles of solute per kilogram of solvent. Since it involves the mass of the solvent (in kilograms), it has units of moles per kilogram (mol/kg).

• Molarity: Molarity is defined as the number of moles of solute per liter of solution. Since it involves the volume of the solution (in liters), it has units of moles per liter (mol/L).

Answer

$(a, d)$

Law of conservation of mass is simply the law of indestructibility of matter during physical or chemical changes. Avogadro law states that equal volumes of different gases contain the same number of molecules under similar conditions of temperature and pressure.

• The Law of definite proportions is related to the statement because it states that a chemical compound always contains exactly the same proportion of elements by mass.
• The Law of multiple proportions is related to the statement because it states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers.

Answer

The mass of a carbon-12 atom was determined by a mass spectrometer and found to be equal to $1.992648 \times 10^{-23} \mathrm{~g}$. It is known that 1 mole of $\mathrm{C}-12$ atom weighing $12 \mathrm{~g}$ contains $\mathrm{N}_{\mathrm{A}}$ number of atoms. Thus,

1 mole of $\mathrm{C}-12$ atoms $=12 \mathrm{~g}=6.022 \times 10^{23}$ atoms

$\Rightarrow \quad 6.022 \times 10^{23}$ atoms of $\mathrm{C}-12$ have mass $=12 \mathrm{~g}$

$\therefore \quad 1$ atom of $\mathrm{C}-12$ will have mass $=\frac{12}{6.022 \times 10^{23}} \mathrm{~g}$

$$ =1.992648 \times 10^{-23} \mathrm{~g} \approx 1.99 \times 10^{-23} \mathrm{~g} $$

Thinking Process

(i) To answer the given calculations, least precise term decide the significant figures.

(ii) To round up a number, left the last digit as such, if the digit next to it is less than 5 and increase it by 1 , if the next digit is greater than 5.

Answer

Least precise term 2.5 or 3.5 has two significant figures.

Hence, the answer should have two significant figures

$$ \frac{2.5 \times 1.25 \times 3.5}{2.01} \approx 5.4415=5.4 $$

Answer

Symbol for SI unit of mole is mol.

One mole is defined as the amount of a substance that contains as many particles and there are atoms in exactly $12 \mathrm{~g}(0.012 \mathrm{~kg})$ of the ${ }^{12} \mathrm{C}$ - isotope.

$$ \frac{1}{12} \mathrm{~g} \text { of }{ }^{12} \mathrm{C} \text {-isotope }=1 \text { mole } $$

Answer

Molality It is defined as the number of moles of solute dissolved in $1 \mathrm{~kg}$ of solvent. It is independent of temperature.

Molarity It is defined as the number of moles of solute dissolved in $1 \mathrm{~L}$ of solution. It depends upon temperature (because, volume of solution $\propto$ temperature).

Thinking Process

To calculate the mass per cent of atom, using the formula

Mass per cent of an element

$$ =\frac{\text { Atomic mass of the element present in the compound }}{\text { Molar mass of the compound }} \times 100 $$

Answer

Mass per cent of calcium $\frac{3 \times \text{(atomic mass of calcium)}}{\text{molecular mass of} Ca_3(PO_4)_2} \times 100$

$ =\frac{120 u}{310 u} \times 100=38.71 $%

$ \text { Mass per cent of phosphorus }=\frac{2 \times(\text { atomic mass of phosphorus })}{\text { molecular mass of } Ca_3 \left( PO_{4}\right)_{2}} \times 100 $

$ =\frac{2 \times 31 u}{310 u} \times 100=20 $%

$ \text { Mass per cent of oxygen }=\frac{8 \times(\text { atomic mass of oxygen })}{\text { molecular mass of } Ca_{3}\left(PO_{4}\right)_{2}} \times 100 $

$ =\frac{8 \times 16 u}{310 u} \times 100=41.29 $%

Answer

For the reaction,

$\underset{2 ~V}2N_{2}(g)+ \underset{1 ~V}{O_{2}(g)} \longrightarrow \underset{2 ~V}2N_2 O(g) $

$ \frac{45.4}{22.7}=2 \quad \frac{22.7}{22.7}=1 \quad \frac{45.4}{22.7}=2 $

Hence, the ratio between the volumes of the reactants and the product in the given question is simple i.e., $2: 1: 2$. It proves the Gay-Lussac’s law of gaseous volumes.

Note Gay-Lussac’s law of gaseous volumes, when gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Answer

(a) Yes, the given statement is true.

(b) According to the law of multiple proportions

(c) $\underset{2 g}{H_{2}}+\underset{16 {~g}}{O_{2}} \longrightarrow \underset{18 {~g}}{H_{2} {O}}$

$\underset{2 g}{H_{2}}+\underset{32 g}{O_{2}} \longrightarrow \underset{34 g}{H_{2} O_{2}}$

Here, masses of oxygen, (i.e., $16 ~g$ in $H_{2} {O}$ and $32 {~g} in_{2} O_{2}$ ) which combine with fixed mass of hydrogen $(2 {~g})$ are in the simple ratio $i . e ., 16: 32$ or $1: 2$.

Answer

Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence), the average atomic mass of the element can be calculated as

$\frac{(\text{ Natural abundance of }^{2} {H} \times \text{ molar mass of }^{2} {H})}{100} $

$\text{ Average atomic mass } =\frac{(\text{ Natural abundance of }^{2} {H} \times \text{ molar mass of }^{2} {H})}{100} $

$=\frac{99.985 \times 1+0.015 \times 2}{100} $

$ =\frac{99.985+0.030}{100}=\frac{100.015}{100}=1.00015 {u}$

Answer

Given that, Mass of $\mathrm{Zn}=32.65 \mathrm{~g}$

1 mole of gas occupies $=22.7 \mathrm{~L}$ volume at STP

Atomic mass of $\quad \mathrm{Zn}=65.3 \mathrm{u}$

The given equation is

$$ \underset{65.3 ~g}{Zn}+2 {HCl} \longrightarrow ZnCl_{2}+\underset{1 ~mol=22.7 {~L} \text{ at STP }}{H_{2}} $$

From the above equation, it is clear that $65.3 \mathrm{~g} \mathrm{Zn}$, when reacts with $\mathrm{HCl}$, produces $=22.7 \mathrm{~L}$ of $\mathrm{H}_{2}$ at STP

$\therefore 32.65 \mathrm{~g} \mathrm{Zn}$, when reacts with $\mathrm{HCl}$, will produce $=\frac{22.7 \times 32.65}{65.3}=11.35 \mathrm{~L}$ of $\mathrm{H}_{2}$ at STP.

Thinking Process

Determine the mass of solution from the given molality of the solution followed by volume of solution relating mass and density to each other, i.e.,

$$ \text { Volume }=\frac{\text { Mass }}{\text { Density }} $$

Then, calculate the molarity of solution as

$$ \text { Molarity }=\frac{\text { Number of moles }}{\text { Volume in litres }} $$

Answer

3 molal solution of $\mathrm{NaOH}$ means 3 moles of $\mathrm{NaOH}$ are dissolved in $1 \mathrm{~kg}$ solvent. So, the mass of solution $=1000 \mathrm{~g}$ solvent $+120 \mathrm{~g} \mathrm{NaOH}=1120 \mathrm{~g}$ solution

(Molar mass of $\mathrm{NaOH}=23+16+1=40 \mathrm{~g}$ and 3 moles of $\mathrm{NaOH}=3 \times 40=120 \mathrm{~g}$ )

$$ \begin{aligned} & \text { Volume of solution }=\frac{\text { Mass of solution }}{\text { Density of solution }} \quad \left(\mathrm{} d=\frac{m}{V}\right)\\ & \qquad \begin{aligned} V & =\frac{1120 \mathrm{~g}}{1.110 \mathrm{~g} \mathrm{~mL}^{-1}}=1009 \mathrm{~mL} \\ \text { Molarity } & =\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})} \\ & =\frac{3 \times 1000}{1009}=2.973 \mathrm{M} \approx 3 \mathrm{M} \end{aligned} \end{aligned} $$

Answer

No, molality of solution does not change with temperature since mass remains unaffected with temperature.

$$ \text { Molality, } m=\frac{\text { moles of solute }}{\text { weight of solvent (in g) }} \times 1000 $$

Thinking Process

(i) To proceed the calculation, first calculate the number of moles of $\mathrm{NaOH}$ and $\mathrm{H}_{2} \mathrm{O}$

(ii) Then, find mole fraction of $\mathrm{NaOH}$ and $\mathrm{H}_{2} \mathrm{O}$ by using the formula,

$$ X_{\mathrm{NaOH}}=\frac{n_{\mathrm{NaOH}}}{n_{\mathrm{NaOH}}+n_{\mathrm{H}_{2} \mathrm{O}}} $$

$$ \left(\operatorname{or} X_{H_{2} {O}}=\frac{n_{H_{2} {O}}}{n_{{NaOH}}+n_{H_{2} {O}}}\right) $$

(iii) Then, calculate molarity $=\frac{W \times 1000}{m \times V}$, so in order to calculate molarity we require volume of solution which is, $V=\frac{m}{\text { specific gravity }}$.

Answer

Number of moles of $\mathrm{NaOH}$,

$n_{NaOH}=\frac{4}{40}=0.1 {~mol} \quad \quad$ $ \lbrace Qn = \frac{Mass(g)}{\text{Molar Mass} (g mol^{-1})} \rbrace$

$Similarly, \quad n_{NaOH}=\frac{36}{18}=2 {~mol} $

$\text { Mole fraction of } {NaOH}, X_{{NaOH}}=\frac{\text { moles of } {NaOH}}{\text { moles of } {NaOH}+\text { moles of } H_{2} {O}} $

$X_{NaOH}=\frac{0.1}{0.1+2}=0.0476 $

$\text{Similarly,} X_{H_{2} {O}}=\frac{n_{H_{2} {O}}}{n_{NaOH}+n_{H_{2} {O}}} $

$=\frac{2}{0.1+2}=0.9524 $

Total mass of solution $=$ mass of solute + mass of solvent

$$ \begin{aligned} & =4+36=40 \mathrm{~g} \\ \text { Volume of solution } & =\frac{\text { Mass of solution }}{\text { specific gravity }}=\frac{40 \mathrm{~g}}{1 \mathrm{~g} \mathrm{~mL}^{-1}}=40 \mathrm{~mL} \\ \text { Molarity } & =\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})} \\ & =\frac{0.1 \times 1000}{40}=2.5 \mathrm{M} \end{aligned} $$

Answer

$$ 2 A+4 B \longrightarrow 3 C+4 D $$

According to the given reaction, 2 moles of $A$ react with 4 moles of $B$.

Hence, 5 moles of $A$ will react with 10 moles of $B\left(\frac{5 \times 4}{2}=10\right.$ moles $)$

(a) It indicates that reactant $B$ is limiting reagent as it will consume first in the reaction because we have only 6 moles of $B$.

(b) Limiting reagent decide the amount of product produced.

According to the reaction,

4 moles of $B$ produces 3 moles of $C$

$\therefore 6$ moles of $B$ will produce $\frac{3 \times 6}{4}=4.5$ moles of $C$.

Note Limiting reagent limits the amount of product formed because it is present in lesser amount and gets consumed first.

Answer

A. $\rightarrow(2)$

B. $\rightarrow$ (3)

C. $\rightarrow(1)$

D. $\rightarrow$ (5)

E. $\rightarrow(4)$

A. Number of moles of $CO_{2}$ molecule $=\frac{\text { Weight in gram of } CO_{2}}{\text { Molecular weight of } CO_{2}}=\frac{88}{44}=2 {~mol}$

B. 1 mole of a substance $=N_{A}$ molecules $=6.022 \times 10^{23}$ molecules

$=$ Avogadro number

$=6.022 \times 10^{23}$ molecules of $H_{2} {O}=1 {~mol}$

C. $22.4 {~L}$ of $O_{2}$ at ${STP}=1 {~mol}$

$5.6 \mathrm{~L}$ of $\mathrm{O}_{2}$ at $\mathrm{STP}=\frac{5.6}{22.4} \mathrm{~mol}=0.25 \mathrm{~mol}$

D. Number of moles of $96 \mathrm{~g}^{\text {of }} \mathrm{O}_{2}=\frac{96}{32} \mathrm{~mol}=3 \mathrm{~mol}$

E. 1 mole of any gas = Avogadro number $=6.022 \times 10^{23}$ molecules

Answer

A. $\rightarrow$ (5)

B. $\rightarrow$ (4)

C. $\rightarrow(2)$

D. $\rightarrow(7)$

E. $\rightarrow(3)$

F. $\rightarrow(6)$

G. $\rightarrow(1)$

H. $\rightarrow(9)$

A. Molarity $=$ concentration in $\mathrm{mol} \mathrm{L}^{-1}$

$$ \text { Molarity }=\frac{\text { Number of moles }}{\text { Volume in litres }} $$

B. Mole fraction = Unitless

C. Mole $=\frac{\text { Mass }(\mathrm{g})}{\text { Molar mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}=\mathrm{mol}$

D. Molality $=$ concentration in mol per $\mathrm{kg}$ solvent

Molality $=\frac{\text { Number of moles }}{\text { Mass of solvent }(\mathrm{kq})}$

$\mathrm{E}$. The $\mathrm{SI}$ unit for pressure is the pascal $(\mathrm{Pa})$, equal to one newton per square metre $\left(\mathrm{N} / \mathrm{m}^{2}\right.$ or $\left.\mathrm{kg} \cdot \mathrm{m}^{-1} \mathrm{~s}^{-2}\right)$. This special name for the unit was added in 1971; before that, pressure in $\mathrm{SI}$ was expressed simply as $\mathrm{N} / \mathrm{m}^{2}$.

F. Unit of luminous intensity = candela.

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $540 \times 10^{12}$ hertz and that has a radiant intensity in that direction of $1 / 683$ watt per steradian.

G. Density $=\frac{\text { mass }}{\text { volume }}=\mathrm{g} \mathrm{mL}^{-1}$

$H$. Unit of mass = kilogram

The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram

Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason ( $\mathrm{R}$ ) is given. Choose the correct option out of the choices given below in each question.

Answer

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

The molecular formula of ethene is $C_{2} H_{4}$ and its empirical formula is $CH_{2}$.

Thus, Molecular formula $=$ Empirical formula $\times 2$

Answer

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Atomic masses of the elements obtained by scientists by comparing with the mass of carbon comes out to be close to whole number value.

Answer

(c) Assertion is true but Reason is false.

0.200 contains 3 while 200 contains only one significant figure because zero at the end or right of a number are significant provided they are on the right side of the decimal point.

Answer

(c) Assertion is false but Reason is true.

Combustion of $16 \mathrm{~g}$ of methane gives $36 \mathrm{~g}$ of water.

$\underset{=16g}{\underset{ 1 mol }{CH_{4}}}+{2O_{2}} \longrightarrow CO_2 + \underset{=36g}{\underset{2 {mol}}{2H_{2} {O}}}$

Answer

(a) $p_{1}=1 \mathrm{~atm}, p_{2}=\frac{1}{2}=0.5 \mathrm{~atm}, T_{1}=273.15, V_{2}=$ ? , $V_{1}=$ ?

$32 \mathrm{~g}$ dioxygen occupies $=22.4 \mathrm{~L}$ volume at STP

$\therefore \quad 1.6 \mathrm{~g}$ dioxygen will occupy $=\frac{22.4 \mathrm{~L} \times 1.6 \mathrm{~g}}{32 \mathrm{~g}}=1.12 \mathrm{~L}$

$$ V_{1}=1.12 \mathrm{~L} $$

From Boyle’s law (as temperature is constant),

$$ \begin{aligned} p_{1} V_{1} & =p_{2} V_{2} \\ V_{2} & =\frac{p_{1} V_{1}}{p_{2}} \end{aligned} $$

$$ =\frac{1 \mathrm{~atm} \times 1.12 \mathrm{~L}}{0.5 \mathrm{~atm}}=2.24 \mathrm{~L} $$

(b) Number of moles of dioxygen $=\frac{\text { Mass of dioxygen }}{\text { Molar mass of dioxygen }}$

$$ \begin{aligned} n_{O_{2}} & =\frac{1.6}{32}=0.05 {~mol} \\ 1 {~mol} \text { of dioxygen contains } & =6.022 \times 10^{23} \text { molecules of dioxygen } \\ \therefore \quad 0.05 \text { mol of dioxygen } & =6.022 \times 10^{23} \times 0.05 \text { molecule of } O_{2} \\ & =0.3011 \times 10^{23} \text { molecules } \\ & =3.011 \times 10^{22} \text { molecules } \end{aligned} $$

Answer

Molar mass of $\mathrm{CaCO}_{3}=40+12+3 \times 16=100 \mathrm{~g} \mathrm{~mol}^{-1}$

Moles of $CaCO_{3}$ in $1000 {~g}, n_{CaCO_{3}}=\frac{\text { Mass }({g})}{\text { Molar mass }}$

$$ \begin{aligned} n_{\mathrm{CaCO}_{3}} & =\frac{1000 \mathrm{~g}}{100 \mathrm{~g} \mathrm{~mol}^{-1}}=10 \mathrm{~mol} \\ \text { Molarity } & =\frac{\text { Moles of solute }(\mathrm{HCl}) \times 1000}{\text { Volume of solution }} \end{aligned} $$

(It is given that moles of $\mathrm{HCl}$ in $250 \mathrm{~mL}$ of $0.76 \mathrm{M} \mathrm{HCl}=n_{\mathrm{HCl}}$ )

$$ 0.76=\frac{n_{\mathrm{HCl}} \times 1000}{250} $$

$$ \begin{aligned} & n_{HCl}=\frac{0.76 \times 250}{1000}=0.19 {~mol} . \\ \end{aligned} $$

$$ CaCO_{3}(s)+2 {HCl}(a q) \longrightarrow CaCl_{2}(a q)+CO_{2}(g)+H_{2} {O}(l) $$

According to the equation,

1 mole of $\mathrm{CaCO}_{3}$ reacts with 2 moles $\mathrm{HCl}$

$\therefore \quad 10$ moles of $\mathrm{CaCO}_{3}$ will react with $\frac{10 \times 2}{1}=20$ moles $\mathrm{HCl}$.

But we have only 0.19 moles $\mathrm{HCl}, \mathrm{so} \mathrm{HCl}$ is limiting reagent and it limits the yield of $\mathrm{CaCl}_{2}$.

Since,

2 moles of $\mathrm{HCl}$ produces 1 mole of $\mathrm{CaCl}_{2}$

0.19 mole of $\mathrm{HCl}$ will produce $\frac{1 \times 0.19}{2}=0.095 \mathrm{~mol} \mathrm{CaCl}_{2}$

Molar mass of $\mathrm{CaCl}_{2}=40+(2 \times 35.5)=111 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore \quad 0.095$ mole of $\mathrm{CaCl}_{2}=0.095 \times 111=10.54 \mathrm{~g}$

Answer

‘Law of multiple proportions’ was first studied by Dalton in 1803 which may be defined as follows

When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another.

e.g., hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.

$\underset{2g}{Hydrogen} + \underset{16g}{Oxygen} \rightarrow \underset{18g}{Water}$

$\underset{2g}{Hydrogen} + \underset{32g}{Oxygen} \rightarrow \underset{34g}{\text{Hydrogen peroxide}}$

Here, the masses of oxygen (i.e., $16 \mathrm{~g}$ and $32 \mathrm{~g}$ ) which combine with a fixed mass of hydrogen $(2 \mathrm{~g})$ bear a simple ratio, i.e., $16: 32$ or $1: 2$.

As we know that, when compounds mixed in different proportionation, Then they form different compounds. In the above examples, when hydrogen is mixed with different proportion of oxygen, then they form water or hydrogen peroxide.

It shows that there are constituents which combine in a definite proportion. These constituents may be atoms. Thus, the law of multiple proportions shows the existence of atoms which combine into molecules.

Thinking Process

In this question, it is seen that the masses of $B$ which combine with the fixed mass of $A$ in different combinations are related to each other by simple whole numbers.

Answer

Mass of $B$ which is combined with fixed mass of $A$ (say $1 \mathrm{~g}$ ) will be $2.5 \mathrm{~g}, 5 \mathrm{~g}, 1.25 \mathrm{~g}$ and $3.75 \mathrm{~g}$. They are in the ratio $2: 4: 1: 3$ which is a simple whole number ratio. Hence, the law of multiple proportions is applicable.