Molality Questions

When the solvent is water and the concentration of the solute is low (i.e., dilute solution), molality and molarity are roughly equal. The approximation, however, fails as a solution becomes more concentrated, involves a solvent other than water, or undergoes temperature changes that could change the density of the solvent. Since the mass of solute and solvent in a solution does not change in these cases, molality is the preferred method of expressing concentration.

Molality in Chemistry Questions with Solutions

Q1. Select the correct relation between molarity and molality.

[m = molecular weight of solute, d = density g/mL]

1. \(\begin{array}{l}M\left ( \frac{m}{100}+\frac{1}{M^{‘}} \right )=d\end{array} \)
2. \(\begin{array}{l}M^{‘}=\left ( \frac{1000 M}{1000d – Mm} \right )\end{array} \)
3. Both (a) and (b)
4. None of these

Correct Answer. (c) Both (a) and (b)

Q2. The density of 1 M solution of HCl is 1.0585 g/mL. The molality of the solution is:

1. 1.0585
2. 1
3. 0.10
4. 0.0585

Correct Answer. (b) 1

Explanation: Molality,

Q3. The mole fraction of a solute in an aqueous solution is 0.2. The molality of the solution will be:

1. 13.88
2. 1.388
3. 0.138
4. 0.0138

Correct Answer. (a) 13.88

Explanation: Given X_B = 0.2, X_A = 0.8

Molality,

Q4. measure the molality of 34.5g of sugar dissolved in 215g of water.

1. 0.559 m
2. 0.613 m
3. 0.603 m
4. 0.554 m

Correct Answer. (d) 0.554 m

Q5. Which of the following is the correct formula for molality?

1. Molality = kilograms of solute ÷ litres of solvent
2. Molality = moles of solute ÷ kilograms of solvent
3. Molality = kilograms of solute ÷ kilograms of solution
4. Molality = moles of solute ÷ moles of solution

Correct Answer. (b) Molality = moles of solute ÷ kilograms of solvent

Q6. Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL of pure water.

Answer.

25.0 g/119.0 g/mol = 0.210 mol

0.210 mol/0.750 kg = 0.280 m

Q7. State true or False. Molality is temperature-dependent.

Answer. False

Q8. Is molality affected by dilution?

Answer. Since molarity and normality are volume dependent, they change with dilution, whereas molality does not.

Q9. Is molarity greater than molality?

Answer. The solution of molality is always greater than molarity. This is because molarity is calculated as mol per unit L (volume of solution), whereas molality is calculated as moles per unit Kg (i.e. mass of solvent). So, in any case, the mass of the solvent is less than the volume of the solution, and thus the molality is greater.

Q10. How can you convert from molarity to molality?

Answer. If you know the density of the solution, you can convert molarity to molality. Calculate the moles of chemical in 1 litre (L) of solution. Because molarity is defined as the number of moles of chemical per litre, this value will simply equal the molarity of the solution.

Q11. How do molality and molarity differ?

Answer. Both molarity and molality are concentration units.

Molarity

The total number of moles of solute per litre of solution is defined as molarity for a given solution. The molarity of a solution is determined by several physical factors, including pressure, temperature, and mass.

Molality

The term molality is also used to describe the concentration of a solution. Molality (m) is defined as the number of moles of solute in one kilogram of solvent. Molality is expressed in the SI unit as moles per kilogram (mol/kg).

Q12. A solution of H₂SO₄ with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution?

Answer.

8.010 m = 8.010 mol / 1 kg of solvent

(8.010 mol) × (98.0768 g/mol) = 785.6 g of solute

785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution.

1785.6 g / 1.354 g/mL = 1318.76 mL

8.01 moles / 1.31876 L = 6.0739 M = 6.074 M.

Q13. What is the molality of a 3.75 M H₂SO₄ solution with a density of 1.230 g/mL?

Answer.

Mass of 1.00 L of solution:

(1000 mL) × (1.230 g/mL) = 1230 g

Mass of 3.75 mol of H₂SO₄ :

(3.75 mol) × (98.0768 g/mol) = 367.788 g

Mass of solvent:

1230 − 367.788 = 862.212 g

Molality:

3.75 mol / 0.862212 kg = 4.35 m

Q14. What is the molality of a solution consisting of 1.34 mL of carbon tetrachloride (CCl₄, density= 1.59 g/mL) in 65.0 mL of methylene chloride (CH₂Cl₂, density = 1.33 g/mL)?

Answer.

Moles of CCl₄:

(1.34 mL) × (1.59 g/mL) = 2.1306 g

2.1306 g / 153.823 g/mol = 0.013851 mol

Mass of the methylene chloride:

(65.0 mL) × (1.33 g/mL) = 86.45 g = 0.08645 kg

Molality:

0.013851 mol / 0.08645 kg = 0.160 m.

Q15. Calculate the molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.

Answer.

Molar mass of C₂H₄O₂ = 12 × 2 + 1 × 4 + 16 × 2 = 60g mol^–1.

Moles of C₂H₄O₂ = 2.5g / 60 g mol^–1 = 0.0417 mol

Mass of benzene in kg = 75 g / 1000 g kg^–1 = 75 × 10^–3 kg

Molality of C₂H₄O₂ = 0.0417 mol / 75 × 10^–3 kg = 0.556 mol kg^–1.

Practise Questions on Molality

Q1. An aqueous solution of urea containing 18g urea in 1500 cm³ of the solution has a density equal to 1.052. If the molecular weight of urea is 60, the molality of the solution is:

1. 0.200 m
2. 0.192 m
3. 0.100 m
4. 1.200 m

Q2. Calculate the molality of a solution containing 16.5 g of naphthalene (C₁₀H₈) in 54.3 g benzene (C₆H₆).

Q3. Calculate the molality (m) of a 7.55 kg sample of a solution of the solute CH₂Cl₂ (molar mass = 84.93 g/mol) dissolved in the solvent acetone (CH₃COCH₃) if the sample contains 929 g of methylene chloride.

Q4. 560 g of 2 m aqueous solution of urea is mixed with 2480 g of 4 m aqueous solution of urea. What is the molality of the resulting solution?

Q5. What is the molality of NaCl in an aqueous solution in which the mole fraction of NaCl is 0.100?

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