Question

A 0.10 M solution of a monoprotic acid (d = 1.01 g/cm³) is 5% ionized. What is the freezing point of the solution ? The molecular weight of the acid is 300 g/mol and K_f(H₂O) = 1.86^o C/m.

• A. -0.189^oC
• B. -0.194^oC
• C. -0.199^oC
• D. None of these

Answer 1

Answer: (C)

Mass of 1 litre of solution = 1010 g

Mass of solvent $= 1 0 1 0 - 3 0 0 \times \text{0.1} \Rightarrow 9 8 0 g$

$\text{Molality(m)} = \frac{\text{0.1}}{\text{0.98}} \Rightarrow \text{0.102} ;$

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m} \cdot \mathrm{i}=(1+\alpha) \mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$

$\text{Δ} \text{T}_{\text{f}} = \text{1.05} \times \text{1.86} \times \text{0.102} = \text{0.199}^{\text{o}} C ;$

$\text{T}_{\text{f}} = 0 - \text{0.199} = - \text{0.199}^{\text{o}} \text{C}$