Some Basic Concepts Of Chemistry
13. Which of the following statements is correct about the reaction given below:
$ 4 Fe(s)+3 O_2(g) \longrightarrow 2 Fe_2 O_3(g) $
(i) Total mass of iron and oxygen in reactants $=$ total mass of iron and oxygen in product therefore it follows law of conservation of mass.
(ii) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed.
(iii) Amount of $Fe_2 O_3$ can be increased by taking any one of the reactants (iron or oxygen) in excess.
(iv) Amount of $Fe_2 O_3$ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.
Show Answer
Answer: (i)14. Which of the following reactions is not correct according to the law of conservation of mass.
(i) $\quad 2 Mg(s)+O_2$ (g) $\longrightarrow 2 MgO(s)$
(ii) $C_3 H_8(g)+O_2(g) \longrightarrow CO_2(g)+H_2 O(g)$
(iii) $P_4$ (s) $+5 O_2(g) \longrightarrow P_4 O _{10}(s)$
(iv) $CH_4(g)+2 O_2(g) \longrightarrow CO_2(g)+2 H_2 O(g)$
Show Answer
Answer: (ii)15. Which of the following statements indicates that law of multiple proportion is being followed.
(i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2.
(ii) Carbon forms two oxides namely $CO_2$ and $CO$, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.
(iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(iv) At constant temperature and pressure $200 mL$ of hydrogen will combine with $100 mL$ oxygen to produce $200 mL$ of water vapour.
II. Multiple Choice Questions (Type-II)
In the following questions two or more options may be correct.
Show Answer
Answer: (ii)
II. Multiple Choice Questions (Type-II)
16. One mole of oxygen gas at STP is equal to ———.
(i) $6.022 \times 10^{23}$ molecules of oxygen
(ii) $6.022 \times 10^{23}$ atoms of oxygen
(iii) $16 g$ of oxygen
(iv) $32 g$ of oxygen
Show Answer
Answer: (i), (iv)17. Sulphuric acid reacts with sodium hydroxide as follows :
$ H_2 SO_4+2 NaOH \longrightarrow Na_2 SO_4+2 H_2 O $
When $1 L$ of $0.1 M$ sulphuric acid solution is allowed to react with $1 L$ of $0.1 M$ sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is
(i) $0.1 mol L^{-1}$
(ii) $7.10 g$
(iii) $0.025 mol L^{-1}$
(iv) $3.55 g$
Show Answer
Answer: (ii), (iii)18. Which of the following pairs have the same number of atoms?
(i) $16 g$ of $O_2(g)$ and $4 g$ of $H_2(g)$
(ii) $16 g$ of $O_2$ and $44 g$ of $CO_2$
(iii) $28 g$ of $N_2$ and $32 g$ of $O_2$
(iv) $12 g$ of $C(s)$ and $23 g$ of $Na(s)$
Show Answer
Answer: (iii), (iv)19. Which of the following solutions have the same concentration?
(i) $20 g$ of $NaOH$ in $200 mL$ of solution
(ii) $0.5 mol$ of $KCl$ in $200 mL$ of solution
(iii) $40 g$ of $NaOH$ in $100 mL$ of solution
(iv) $20 g$ of $KOH$ in $200 mL$ of solution
Show Answer
Answer: (i), (ii)20. $16 g$ of oxygen has same number of molecules as in
(i) $16 g$ of $CO$
(ii) $28 g$ of $N_2$
(iii) $14 g$ of $N_2$
(iv) $1.0 g$ of $H_2$
Show Answer
Answer: (iii), (iv)21. Which of the following terms are unitless?
(i) Molality
(ii) Molarity
(iii) Mole fraction
(iv) Mass percent
Show Answer
Answer: (iii), (iv)22. One of the statements of Dalton’s atomic theory is given below:
“Compounds are formed when atoms of different elements combine in a fixed ratio”
Which of the following laws is not related to this statement?
(i) Law of conservation of mass
(ii) Law of definite proportions
(iii) Law of multiple proportions
(iv) Avogadro law
III. Short Answer Type
Show Answer
Answer: (i), (iv)
III. Short Answer Type
23. What will be the mass of one atom of $C-12$ in grams?
Show Answer
Answer: $1.992648 \times 10^{-23} g \approx 1.99 \times 10^{-23} g$24. How many significant figures should be present in the answer of the following calculations?
$ \frac{2.5 \times 1.25 \times 3.5}{2.01} $
Show Answer
Answer: 225. What is the symbol for SI unit of mole? How is the mole defined?
Show Answer
Answer: Symbol for SI Unit of mole is mol.
One mole is defined as the amount of a substance that contains as many particles or entities as there are atoms in exactly $12 g(0.012 kg)$ of the ${ }^{12} C$ isotope.
26. What is the difference between molality and molarity?
Show Answer
Answer: Molality is the number of moles of solute present in one kilogram of solvent but molarity is the number of moles of solute dissolved in one litre of solution.
Molality is independent of temperature whereas molarity depends on temperature.
27. Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate $Ca_3(PO_4)_2$.
Show Answer
Answer: Mass percent 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 % $
Mass percent of phosphorus $=\frac{2 \times(\text{ atomic mass of phosphorus })}{\text{ molecular mass of } Ca_3(PO_4)_2} \times 100$
$ =\frac{2 \times 31 u}{310 u} \times 100=20 % $
Mass percent of oxygen $=\frac{8 \times(\text{ Atomic mass of oxygen })}{\text{ molecular mass of } Ca_3(PO_4)_2} \times 100$
$ =\frac{8 \times 16 u}{310 u} \times 100=41.29 % $
28. 45.4 $L$ of dinitrogen reacted with $22.7 L$ of dioxygen and $45.4 L$ of nitrous oxide was formed. The reaction is given below:
$ 2 N_2(g)+O_2(g) \longrightarrow 2 N_2 O(g) $
Which law is being obeyed in this experiment? Write the statement of the law?
5 Some Basic Concepts of Chemistry
Show Answer
Answer: According to Gay Lussac’s law of gaseous volumes, gases combine or are produced in a chemical reaction in a simple ratio by volume, provided that all gases are at the same temperature and pressure.29. If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.
(a) Is this statement true?
(b) If yes, according to which law?
(c) Give one example related to this law.
Show Answer
Answer: (a) Yes
(b) According to the law of multiple proportions
(c) $H_2+O_2 \to H_2 O$
$2 g \quad 16 g \quad 18 g$
(c) $H_2+O_2 \to H_2 O_2$
$2 g \quad 32 g \quad 34 g$
Here masses of oxygen, (i.e., $16 g$ in $H_2 O$ and $32 g^{2}$ in $H_2 O_2$ ) which combine with fixed mass of hydrogen $(2 g)$ are in the simple ratio i.e., $16: 32$ or $1: 2$
{(Natural abundance of ${ }^{1} H \times$ molar mass) +
30. Calculate the average atomic mass of hydrogen using the following data :
| Isotope | % Natural abundance | Molar m |
|---|---|---|
| ${ }^{1} H$ | 99.985 | 1 |
| ${ }^{2} H$ | 0.015 | 2 |
Show Answer
Answer: Average Atomic Mass $=\frac{.(\text{ Natural abundance of }{ }^{2} H \times \text{ molar mass of }{ }^{2} H)}}{100}$
$ \begin{aligned} & =\frac{99.985 \times 1+0.015 \times 2}{100} \\ & =\frac{99.985+0.030}{100}=\frac{100.015}{100}=1.00015 u \end{aligned} $
31. Hydrogen gas is prepared in the laboratory by reacting dilute $HCl$ with granulated zinc. Following reaction takes place.
$ Zn+2 HCl \longrightarrow ZnCl_2+H_2 $
Calculate the volume of hydrogen gas liberated at STP when $32.65 g$ of zinc reacts with $HCl$. $1 mol$ of a gas occupies $22.7 L$ volume at STP; atomic mass of $Zn=65.3 u$.
Show Answer
Answer: From the equation, $63.5 g$ of zinc liberates 22.7 litre of hydrogen. So $32.65 g$ of zinc will liberate
$ 32.65 g Zn \times \frac{22.7 L H_2}{65.3 g Zn^{2}}=\frac{22.7}{2} L=11.35 L $
32. The density of 3 molal solution of $NaOH$ is $1.110 g mL^{-1}$. Calculate the molarity of the solution.
Show Answer
Answer: 3 molal solution of $NaOH$ means that 3 mols of $NaOH$ are dissolved in $1000 g$ of solvent.
$\therefore$ Mass of Solution $=$ Mass of Solvent + Mass of Solute
$ =1000 g+(3 \times 40 g)=1120 g $
Volume of Solution $=\frac{1120}{1.110} mL=1009.00 mL$
(Since density of solution $=1.110 g mL^{-1}$ )
Since $1009 mL$ solution contains $3 mols$ of $NaOH$
$\therefore$ Molarity $=\frac{\text{ Number of moles of solute }}{\text{ Volume of solution in litre }}$
$ =\frac{3 mol}{1009.00} \times 1000=2.97 M $
33. Volume of a solution changes with change in temperature, then, will the molality of the solution be affected by temperature? Give reason for your answer.
Show Answer
Answer: No, Molality of solution does not change with temperature since mass remains unaffected with temperature.34. If $4 g$ of $NaOH$ dissolves in $36 g$ of $H_2 O$, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is $1 g mL^{-1}$ ).
Show Answer
Answer: Mass of $NaOH=4 g$
Number of moles of $NaOH=\frac{4 g}{40 g}=0.1 mol$
Mass of $H_2 O=36 g$
Number of moles of $H_2 O=\frac{36 g}{18 g}=2 mol$
Mole fraction of water $=$ Number of moles of $H_2 O$
No. of moles of water + No. of moles of $NaOH$
$=\frac{2}{2+0.1}=\frac{2}{2.1}=0.95$
Mole fraction of water =$\frac{\text Number of moles of H_2O}{\text No. of moles of water + No. of moles of NaOH}$
$ =\frac{0.1}{2+0.1}=\frac{0.1}{2.1}=0.047 $
Mass of solution $=$ mass of water + mass of $NaOH=36 g+4 g=40 g$
Volume of solution $=40 \times 1=40 mL$
(Since specific gravity of solution is $=1 g mL^{-1}$ )
Molarity of solution $=\frac{\text{ Number of moles of solute }}{\text{ Volume of solution in litre }}$
$ =\frac{0.1 mol NaOH}{0.04 L}=2.5 M $
35. The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2 A+4 B \to 3 C+4 D$, when 5 moles of $A$ react with 6 moles of $B$, then
(i) which is the limiting reagent?
(ii) calculate the amount of $C$ formed?
IV. Matching Type
Show Answer
Answer: $2 A+4 B \to 3 C+4 D$
According to the above equation, 2 mols of ’ $A$ ’ require 4 mols of ’ $B$ ’ for the reaction.
Hence, for $5 mols$ of ’ $A$ ‘, the moles of ’ $B$ ’ required $=5 mol$ of $A \times \frac{4 mol \text{ of } B}{2 mol \text{ of } A}$
$ =10 mol B $
But we have only 6 mols of ’ $B$ ‘, hence, ’ $B$ ’ is the limiting reagent. So amount of ’ $C$ ’ formed is determined by amount of ’ $B$ ‘.
Since 4 mols of ’ $B$ ’ give 3 mols of ’ $C$ ‘. Hence 6 mols of ’ $B$ ’ will give
$ 6 mol \text{ of } B \times \frac{3 mol \text{ of } C}{4 mol \text{ of } B}=4.5 mol \text{ of } C $
IV. Matching Type
36. Match the following:
| (i) $88 g$ of $CO_2$ $\quad$ | (a) $0.25 mol$ |
|---|---|
| (ii) $6.022 \times 10^{23}$ molecules of $H_2 O$ | (b) $2 mol$ |
| (iii) 5.6 litres of $O_2$ at STP $ | (c) $1 mol$ |
| (iv) $96 g$ of $O_2$ | (d) $6.022 \times 10^{23}$ molecules |
| (v) $1 mol$ of any gas | (e) $3 mol$ |
Show Answer
Answer: $(i) \to (b) \quad (ii) \to (c) \quad (iii) \to (a) \quad (iv) \to (e) \quad (v) \to (d)$37. Match the following physical quantities with units
| Physical quantity | Unit |
|---|---|
| (i) Molarity | (a) $g mL^{-1}$ |
| (ii) Mole fraction | (b) mol |
| (iii) Mole | (c) Pascal |
| (iv) Molality | (d) Unitless |
| (v) Pressure | (e) $mol L^{-1}$ |
| (vi) Luminous intensity | (f) Candela |
| (vii) Density | (g) $mol kg^{-1}$ |
| (viii) Mass | (h) $Nm^{-1}$ |
| (i) $kg$ |
V. Assertion and Reason Type
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Show Answer
Answer: $(i) \to (e) \quad (ii) \to (d) \quad (iii) \to (b) \quad (iv) \to (g) \quad (v) \to (c), (h) \quad (vi) \to (f) \quad (vii) \to (a) \quad (viii) \to (i)$
Solution: N/A
VI. Assertion and Reason Type
38. Assertion (A) : The empirical mass of ethene is half of its molecular mass.
Reason (R) : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
(i) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
(ii) $A$ is true but $R$ is false.
(iii) $A$ is false but $R$ is true.
(iv) Both $A$ and $R$ are false.
Show Answer
Answer: (i)39. Assertion (A) : One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.
Reason (R) : Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.
(i) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
Show Answer
Answer: (ii)40. Assertion (A) : Significant figures for 0.200 is 3 where as for 200 it is 1 .
Reason (R) : Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
(i) Both $A$ and $R$ are true and $R$ is correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not a correct explanation of $A$. (iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
41. Assertion (A) : Combustion of $16 g$ of methane gives $18 g$ of water.
Reason (R) : In the combustion of methane, water is one of the products.
(i) Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
(ii) $A$ is true but $R$ is false.
(iii) $A$ is false but $R$ is true.
(iv) Both $A$ and $R$ are false.
VI. Long Answer Type
Show Answer
Answer: (iii)
V. Long Answer Type
42. A vessel contains $1.6 g$ of dioxygen at STP $(273.15 K, 1 atm$ pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate
(i) volume of the new vessel.
(ii) number of molecules of dioxygen.
Show Answer
Answer: (i) $p_1=1 atm$,
$ T_1=273 K, \quad V_1=? $
$32 g$ of oxygen occupies $22.4 L$ of volume at STP*
Hence, $1.6 g$ of oxygen will occupy, $1.6 g$ oxygen $\times \frac{22.4 L}{32 g \text{ oxygen }}=1.12 L$
$V_1=1.12 L$
$p_2=\frac{p_1}{2}=\frac{1}{2}=0.5 atm$.
$V_2=$ ?
According to Boyle’s law :
$ \begin{aligned} & p_1 V_1=p_2 V_2 \\ & V_2=\frac{p_1 \times V_1}{p_2}=\frac{1 atm . \times 1.12 L}{0.5 atm .}=2.24 L \end{aligned} $
- Old STP conditions $273.15 K, 1 atm$, volume occupied by $1 mol$ of gas = $22.4 L$. New STP conditions $273.15 K, 1$ bar, volume occupied by a gas $=22.7 L$. (ii) Number of molecules of oxygen in the vessel $=\frac{6.022 \times 10^{23} \times 1.6}{32}$
$ =3.011 \times 10^{22} $
43. Calcium carbonate reacts with aqueous $HCl$ to give $CaCl_2$ and $CO_2$ according to the reaction given below:
$ CaCO_3(s)+2 HCl(aq) \longrightarrow CaCl_2(aq)+CO_2(g)+H_2 O(l) $
What mass of $CaCl_2$ will be formed when $250 mL$ of $0.76 M HCl$ reacts with $1000 g$ of $CaCO_3$ ? Name the limiting reagent. Calculate the number of moles of $CaCl_2$ formed in the reaction.
Show Answer
Answer: Number of moles of $HCl=250 mL \times \frac{0.76 M}{1000}=0.19 mol$
Mass of $CaCO_3=1000 g$
Number of moles of $CaCO_3=\frac{1000 g}{100 g}=10 mol$
According to given equation $1 mol$ of $CaCO_3$ (s) requires $2 mol$ of $HCl$ (aq). Hence, for the reaction of $10 mol$ of $CaCO_3(s)$ number of moles of $HCl$ required would be:
$10 ~ mol ~ CaCO_3 \times \frac{2 \text ~ mol ~ HCl ~ (aq)}{1 ~ \text mol ~ CaCO_3 ~ (s)}= 20 ~ mol ~ HCl ~ (aq)$
But we have only $0.19 mol HCl(aq)$, hence, $HCl(aq)$ is limiting reagent.
So amount of $CaCl_2$ formed will depend on the amount of $HCl$ available.
Since, $2 mol HCl(aq)$ forms $1 mol$ of $CaCl_2$, therefore, $0.19 mol$ of $HCl(aq)$ would give:
$0.19 \text ~ mol ~ HCl ~ (aq) \times \frac {1 ~ \text mol ~ CaCl_2 ~ (aq)}{ 2 ~ \text mol ~ HCl ~ (aq) = 0.095 ~ \text mol}$
or
$ 0.095 \times \text{ molar mass of } CaCl_2=0.095 \times 111=10.54 g $ Solution: N/A
44. Define the law of multiple proportions. Explain it with two examples. How does this law point to the existance of atoms?
45. A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as $B$, each weighing 5 grams. Consider the combinations $AB, AB_2, A_2 B$ and $A_2 B_3$ and show that law of multiple proportions is applicable.
