Question:

Which one of the following will have largest number of atoms ? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl₂ (g)

Answer:

Step 1: Determine the atomic mass of each element. Au = 197 g/mol, Na = 23 g/mol, Li = 6.94 g/mol, Cl₂ = 70.9 g/mol.

Step 2: Determine the number of moles of each element. 1 g Au (s) = 0.005 moles, 1 g Na (s) = 0.043 moles, 1 g Li (s) = 0.144 moles, 1 g of Cl₂ (g) = 0.014 moles.

Step 3: Determine the number of atoms of each element. 1 g Au (s) = 0.005 moles x 6.02x10²³ atoms/mole = 3.01x10^22 atoms, 1 g Na (s) = 0.043 moles x 6.02x10²³ atoms/mole = 2.59x10²³ atoms, 1 g Li (s) = 0.144 moles x 6.02x10²³ atoms/mole = 8.67x10²² atoms, 1 g of Cl₂ (g) = 0.014 moles x 6.02x10²³ atoms/mole = 8.43x10²² atoms.

Step 4: Compare the number of atoms of each element. The element with the largest number of atoms is Na (s) with 2.59x10²³ atoms.

Question:

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.

Calculate (i) empirical formula, (ii) molar mass of the gas and (iii) molecular formula.

Answer:

(i) Empirical Formula:

1. Calculate the moles of carbon dioxide (CO₂) and water (H₂O) produced from the burning of the welding fuel gas:

Moles of CO2 = 3.38 g/44.01 g/mol = 0.0765 mol Moles of H2O = 0.690 g/18.02 g/mol = 0.0383 mol

2. Calculate the moles of carbon (C) and hydrogen (H) present in the welding fuel gas:

Moles of C = 0.0765 mol/2 = 0.0383 mol Moles of H = 0.0383 mol/2 = 0.0192 mol

3. Calculate the empirical formula of the welding fuel gas:

Empirical Formula = CH₂

(ii) Molar Mass of the Gas:

1. Calculate the molar mass of the welding fuel gas using the empirical formula:

Molar Mass = (1 mol C x 12.01 g/mol) + (2 mol H x 1.01 g/mol) = 14.04 g/mol

(iii) Molecular Formula:

1. Calculate the moles of the welding fuel gas present in 10.0 L (measured at STP):

Moles of Gas = 11.6 g/14.04 g/mol = 0.827 mol

2. Calculate the molar mass of the molecular formula using the moles of the welding fuel gas:

Molar Mass = 0.827 mol x 14.04 g/mol = 11.6 g/mol

3. Calculate the molecular formula of the welding fuel gas:

Molecular Formula = CH₂

Question:

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. [Assume the density of water to be one]

Answer:

Step 1: Calculate the mass of ethanol in the solution.

Mass of ethanol = 0.040 x density of water x volume of solution

Step 2: Calculate the moles of ethanol in the solution.

Moles of ethanol = mass of ethanol / molecular weight of ethanol

Step 3: Calculate the molarity of the solution.

Molarity = moles of ethanol / volume of solution

Question:

In a reaction A+B₂→AB₂, identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B₂ (ii) 2 mol A + 3 mol B₂ (iii) 100 atoms of A + 100 molecules of B₂ (iv) 5 mol A + 2.5 mol B₂ (v) 2.5 mol A + 5 mol B₂

Answer:

(i) 300 atoms of A + 200 molecules of B₂

There are 300 atoms of A and 200 molecules of B₂, which is equivalent to 600 atoms of B₂. Since there are more atoms of B2 than A, B₂ is the limiting reagent.

(ii) 2 mol A + 3 mol B2

There are 2 moles of A and 3 moles of B₂. Since there are more moles of B₂ than A, B₂ is the limiting reagent.

(iii) 100 atoms of A + 100 molecules of B₂

There are 100 atoms of A and 100 molecules of B₂, which is equivalent to 200 atoms of B₂. Since there are equal amounts of A and B₂, neither is the limiting reagent.

(iv) 5 mol A + 2.5 mol B₂

There are 5 moles of A and 2.5 moles of B₂. Since there are more moles of A than B₂, B₂ is the limiting reagent.

(v) 2.5 mol A + 5 mol B₂

There are 2.5 moles of A and 5 moles of B₂. Since there are more moles of B₂ than A, B₂ is the limiting reagent.

Question:

Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808

Answer:

(i) 34.2 (ii) 10.4 (iii) 0.046 (iv) 2810

Question:

How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,00 (v) 500.0 (vi) 2.0034

Answer:

(i) 3 significant figures (ii) 3 significant figures (iii) 4 significant figures (iv) 5 significant figures (v) 4 significant figures (vi) 5 significant figures

Question:

How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Answer:

Step 1: Determine the molecular weight of copper sulphate (CuSO₄).

The molecular weight of copper sulphate (CuSO₄) is 159.609 g/mol.

Step 2: Calculate the amount of copper present in 100 g of copper sulphate (CuSO₄).

The amount of copper present in 100 g of copper sulphate (CuSO₄) is 100 g x (1 mol/159.609 g) = 0.6265 mol.

Step 3: Calculate the amount of copper obtained from 100 g of copper sulphate (CuSO₄).

The amount of copper obtained from 100 g of copper sulphate (CuSO₄) is 0.6265 mol x (63.546 g/mol) = 39.87 g.

Question:

Calculate the molar mass of CO₂.

Answer:

Step 1: Identify the elements in CO₂. Answer: Carbon and Oxygen

Step 2: Look up the atomic masses of the elements. Answer: Carbon = 12.01 g/mol; Oxygen = 16.00 g/mol

Step 3: Calculate the molar mass of CO₂. Answer: Molar mass of CO₂ = 12.01 g/mol + (2 x 16.00 g/mol) = 44.01 g/mol

Question:

Calculate the mass percent of O present in sodium sulphate (Na₂SO₄).

Answer:

Step 1: Calculate the molar mass of sodium sulphate. Molar mass of Na2SO4 = (2 x 23) + (32 + 4 x 16) = 142 g/mol

Step 2: Calculate the mass of oxygen present in sodium sulphate. Mass of oxygen = (4 x 16) = 64 g

Step 3: Calculate the mass percent of oxygen present in sodium sulphate. Mass percent of oxygen = (64/142) x 100 = 45%

Question:

How many significant figures should be present in the answer of the following calculations? (i) 0.02856×298.15×0.112/0.5785 (ii) 5×5.364 (iii) 0.0125+0.7864+0.0215

Answer:

(i) 5 significant figures; (ii) 4 significant figures; (iii) 4 significant figures.

Question:

If the density of methanol is 0.793kgL^-1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Answer:

1. Calculate the mass of 2.5 L of 0.25 M methanol solution: Molarity (M) = moles of solute/liters of solution Moles of solute = Molarity x Liters of solution Moles of solute = 0.25 M x 2.5 L = 0.625 moles Mass of solute = moles x molar mass of solute Molar mass of methanol = 32 g/mol Mass of methanol = 0.625 moles x 32 g/mol = 20 g

2. Calculate the volume of methanol needed to make 2.5 L of 0.25 M methanol solution: Density (ρ) = mass/volume Volume = mass/density Volume = 20 g/0.793 kg/L = 25.2 L

Question:

Convert the following into SI unit: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg

Answer:

(i) 28.7 pm = 2.87 x 10^-10 m

(ii) 15.15 pm = 1.515 x 10^-10 m

(iii) 25365 mg = 2.5365 x 10^-3 kg

Question:

Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer:

(i) When 1 mole of carbon is burnt in air, it produces 1 mole of carbon dioxide.

(ii) When 1 mole of carbon is burnt in 16 g of dioxygen, it produces 1 mole of carbon dioxide.

(iii) When 2 moles of carbon are burnt in 16 g of dioxygen, it produces 2 moles of carbon dioxide.

Question:

Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4

Answer:

(i) Molar mass of H₂O = 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol

(ii) Molar mass of CO₂ = 12.01 g/mol + 2(15.99 g/mol) = 44.01 g/mol

(iii) Molar mass of CH₄ = 1.008 g/mol + 4(12.01 g/mol) = 16.04 g/mol

Question:

If 3 moles of ethane (C₂H₆) are given, calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.

Answer:

(i) Number of moles of carbon atoms: 3 moles of ethane contains 6 moles of carbon atoms. Answer: 6 moles of carbon atoms.

(ii) Number of moles of hydrogen atoms: 3 moles of ethane contains 18 moles of hydrogen atoms. Answer: 18 moles of hydrogen atoms.

(iii) Number of molecules of ethane: 3 moles of ethane contains 3 molecules of ethane. Answer: 3 molecules of ethane.

Question:

Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Answer:

(i) 4.8 x 10^{-3} (ii) 2.34 x 10^{5} (iii) 8.008 x 10^{3} (iv) 5.00 x 10^{2} (v) 6.0012 x 10^{0}

Question:

Calculate the atomic mass (average) of chlorine using the following data : % Natural Abundance ³⁵Cl75.77, Molar Mass 34.9689 and Abundance ³⁷Cl24.23, Molar Mass 36.9659

Answer:

Step 1: Calculate the total mass of chlorine atoms with 35Cl isotope.

Total mass of 35Cl = 75.77% x 34.9689 = 26.48 g

Step 2: Calculate the total mass of chlorine atoms with 37Cl isotope.

Total mass of 37Cl = 24.23% x 36.9659 = 9.05 g

Step 3: Calculate the total mass of chlorine atoms.

Total mass of chlorine atoms = 26.48 g + 9.05 g = 35.53 g

Step 4: Calculate the atomic mass (average) of chlorine.

Atomic mass (average) of chlorine = 35.53 g/2 = 17.765 g

Question:

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively.

Answer:

Step 1: Calculate the molar mass of the oxide. Molar mass = (69.9 g/mol x 55.845 g/mol) + (30.1 g/mol x 16.00 g/mol) = 159.7 g/mol

Step 2: Calculate the empirical formula of the oxide. Empirical formula = (69.9/159.7) Fe + (30.1/159.7) O = 0.437 Fe + 0.188 O

Step 3: Determine the molecular formula of the oxide. Molecular formula = (0.437 Fe + 0.188 O) x 2 = 0.874 Fe + 0.376 O = Fe₂O₃

Question:

Calculate the mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1.

Answer:

Step 1: Calculate the number of moles of sodium acetate required to make 500 ml of 0.375 molar aqueous solution.

Number of moles = 0.375 moles/L × 500 ml = 0.188 moles

Step 2: Calculate the mass of sodium acetate required to make 500 ml of 0.375 molar aqueous solution.

Mass of sodium acetate = 0.188 moles × 82.0245 g/mol = 15.35 g

Question:

A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass. (ii) Determine the molality of chloroform in the water sample.

Answer:

(i) 15 ppm (by mass) = 0.0015% (by mass)

(ii) Molality = (mass of solute (g))/(mass of solvent (kg))

Molality = (15 ppm (by mass) × 0.001 g/1 ppm) / (1 kg)

Molality = 0.0000015 kg/kg = 1.5 x 10^-5 mol/kg

Question:

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N₂(g)+H₂(g)→2NH₃(g) (i) Calculate the mass of ammonia produced if 2.00×10³ g dinitrogen reacts with 1.00×10³ g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?

Answer:

(i) The mass of ammonia produced can be calculated using the following equation:

Mass of ammonia = 2 x (mass of dinitrogen) x (mass of dihydrogen)

= 2 x (2.00 x 10³ g) x (1.00 x 10³ g)

= 4.00 x 10⁶ g

(ii) No, none of the two reactants will remain unreacted as the equation is a balanced equation.

(iii) No, none of the reactants will remain unreacted.

Question:

What is the SI unit of mass? How is it defined?

Answer:

Answer: The SI unit of mass is the kilogram (kg). It is defined as the mass of the international prototype of the kilogram, which is a cylinder made of a platinum-iridium alloy stored in a vault at the International Bureau of Weights and Measures in France.

Question:

Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41gmL^-1 and the mass per cent of nitric acid in it being 69%.

Answer:

Step 1: Calculate the mass of nitric acid in the sample. Mass of nitric acid = (69/100) x (1.41 g/mL)

Step 2: Calculate the moles of nitric acid in the sample. Moles of nitric acid = (Mass of nitric acid)/(Molar mass of nitric acid)

Step 3: Calculate the concentration of nitric acid in moles per litre. Concentration of nitric acid = (Moles of nitric acid)/(Volume of sample in litres)

Question:

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Answer:

1. Calculate the molar mass of the oxide of iron: Molar mass of iron = 55.85 g/mol Molar mass of dioxygen = 32.00 g/mol

Molar mass of oxide of iron = 55.85 g/mol + 32.00 g/mol = 87.85 g/mol

2. Calculate the mass of each element in the oxide of iron: Mass of iron = 69.9% x 87.85 g/mol = 61.44 g/mol Mass of dioxygen = 30.1% x 87.85 g/mol = 26.41 g/mol

3. Calculate the moles of each element in the oxide of iron: Moles of iron = 61.44 g/mol ÷ 55.85 g/mol = 1.10 moles Moles of dioxygen = 26.41 g/mol ÷ 32.00 g/mol = 0.82 moles

4. Calculate the simplest whole number ratio of the elements in the oxide of iron: Ratio of iron to dioxygen = 1.10 moles ÷ 0.82 moles = 1.34

5. Determine the empirical formula of the oxide of iron: Empirical formula = Fe1.34O0.82

Question:

What is the concentration of sugar (C₁₂H₂₂O₁₁) in molL_-1 if its 20 g are dissolved in enough water to make a final volume up to 2L?

Answer:

Step 1: Calculate the molar mass of sugar (C₁₂H₂₂O₁₁). Molar mass of sugar = 12(12.01 g/mol) + 22(1.008 g/mol) + 11(15.999 g/mol) = 342.29 g/mol

Step 2: Calculate the number of moles of sugar. Number of moles of sugar = 20 g/ 342.29 g/mol = 0.0584 mol

Step 3: Calculate the concentration of sugar in molL^-1. Concentration of sugar = 0.0584 mol/2L = 0.0292 molL^-1

Question:

Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa=1Nm^-2 If mass of air at sea level is 1034 gcm^-2, calculate the pressure in pascal.

Answer:

Step 1: Convert the mass of air to Newton. 1 gcm^-2 = 9806.65 Nm^-2

Therefore, 1034 gcm^-2 = 1034 x 9806.65 Nm^-2 = 1013907.1 Nm^-2

Step 2: Calculate the pressure in pascal.

Pressure = Force/Area

Therefore, Pressure = 1013907.1 Nm^-2/1m²

Pressure = 1013907.1 Pa

Question:

What will be the mass of one 12C atom in g ?

Answer:

Step 1: Determine the atomic mass of 12C The atomic mass of 12C is 12.

Step 2: Convert the atomic mass to grams 1 atomic mass unit (amu) is equal to 1.66 x 10^-24 g.

Therefore, 12 amu is equal to 19.92 x 10^-24 g.

Step 3: Calculate the mass of one 12C atom The mass of one 12C atom is 19.92 x 10^-24 g.

Question:

What do you mean by significant figures?

Answer:

Significant figures refer to the number of digits in a value that are known with some degree of certainty. Significant figures are used to provide an indication of the precision of a measurement or calculation. They are also used to express the accuracy of a measurement or calculation. For example, a value of 1.2345 has five significant figures, while a value of 1.2 has three significant figures.

Question:

If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer:

1. Dihydrogen gas (H₂) + Dioxygen gas (O₂) → Water vapour (H₂O)

2. 10 volumes of H₂ + 5 volumes of O₂ → x volumes of H₂O

3. 10 volumes of H₂ + 5 volumes of O₂ → 10 volumes of H₂O

Question:

How are 0.50 m Na₂CO₃ and 0.50 M Na₂CO₃ different?

Answer:

1. 0.50 m Na₂CO₃ is a concentration expressed in molarity, which is a measure of the number of moles of solute per liter of solution.

2. 0.50 M Na₂CO₃ is a concentration expressed in molality, which is a measure of the number of moles of solute per kilogram of solvent.

Question:

Calculate the number of atoms in each of the following: (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He

Answer:

(i) 52 moles of Ar Answer: 3.2 x 10²⁴ atoms

(ii) 52 u of He Answer: 3.2 x 10²² atoms

(iii) 52 g of He Answer: 8.4 x 10²³ atoms

Question:

Calculate the molar mass of CH4.

Answer:

Step 1: Identify the elements present in CH4. Answer: Carbon (C) and Hydrogen (H)

Step 2: Find the atomic mass of each element. Answer: Carbon (C): 12.0107 g/mol; Hydrogen (H): 1.00794 g/mol

Step 3: Multiply the atomic mass of each element by the number of atoms present in CH4. Answer: Carbon (C): 12.0107 g/mol x 1 atom = 12.0107 g/mol; Hydrogen (H): 1.00794 g/mol x 4 atoms = 4.03176 g/mol

Step 4: Add the two results together to calculate the molar mass of CH4. Answer: 12.0107 g/mol + 4.03176 g/mol = 16.04246 g/mol

Question:

If the speed of light is 3.0×108ms^-1, calculate the distance covered by light in 2.00 ns.

Answer:

Step 1: Convert 2.00 ns to seconds.

1 ns = 10^-9 s

2.00 ns = 2.00 x 10^-9 s

Step 2: Calculate the distance covered by light.

Distance = Speed x Time

Distance = 3.0 x 10⁸ m/s x 2.00 x 10^-9 s

Distance = 6.00 x 10^-1 m

Question:

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction: 4HCl(aq)+MnO₂(s)→2H₂O(I)+MnCl₂(aq)+Cl₂(g). How many grams of HCl react with 5.0 g of manganese dioxide ?

Answer:

1. Calculate the moles of manganese dioxide (MnO2): MnO₂ = 5.0 g Molar Mass of MnO₂ = 86.94 g/mol

Moles of MnO₂ = 5.0 g / 86.94 g/mol = 0.0575 mol

2. Calculate the moles of hydrochloric acid (HCl): The equation shows that 4 moles of HCl are needed for every 1 mole of MnO2.

Moles of HCl = 4 x 0.0575 mol = 0.23 mol

3. Calculate the grams of hydrochloric acid (HCl): Molar Mass of HCl = 36.5 g/mol

Grams of HCl = 0.23 mol x 36.5 g/mol = 8.395 g

Question:

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction,
CaCO₃(s)+2HCl(aq)→CaCl₂(aq)+CO₂(g)+H₂O(l). What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl ?

Answer:

1. Calculate the moles of HCl: Moles of HCl = 0.75 M * 0.025 L = 0.01875 moles

2. Calculate the moles of CaCO3 required: Moles of CaCO₃ = 2 x 0.01875 moles = 0.0375 moles

3. Calculate the mass of CaCO₃ required: Mass of CaCO₃ = 0.0375 moles x 100.09 g/mol = 3.75 g