Question

When '$x \times 10^{-2} \, \text{mL}$' methanol (molar mass $= 32 \, \text{g}$; density $= 0.792 \, \text{g/cm}^3$) is added to $100 \, \text{mL}$ water (density $= 1 \, \text{g/cm}^3$), the following diagram is obtained.

x = __________ (nearest integer)  
[Given: Molal freezing point depression constant of water at $273.15 \, \text{K}$ is $1.86 \, \text{K kg mol}^{-1}$]

Answer 1

Correct Answer: 543

• Determine freezing point depression (\( \Delta T_f \)):

\( \Delta T_f = 273.15 \, \text{K} - 270.65 \, \text{K} = 2.5 \, \text{K} \)

• Apply the freezing point depression formula:

\( \Delta T_f = K_f \cdot m \implies 2.5 \, \text{K} = 1.86 \, \text{K} \cdot \text{kg/mol} \times \frac{n \, \text{mol}}{0.1 \, \text{kg}} \)

• Isolate moles of methanol (\( n \)):

\( n = \frac{2.5 \, \text{K}}{1.86 \, \text{K} \cdot \text{kg/mol} \times \frac{1}{0.1 \, \text{kg}}} = 0.1344 \, \text{mol} \)

• Calculate mass of methanol (\( w \)):

\( w = n \times \text{Molar Mass} = 0.1344 \, \text{mol} \times 32 \, \text{g/mol} = 4.3 \, \text{g} \)

• Calculate volume of methanol:

\( \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{4.3 \, \text{g}}{0.792 \, \text{g/mL}} = 5.43 \, \text{mL} = 5.43 \times 10^{0} \, \text{mL} \)

Answer: \( x = 5.43 \times 10^{0} \)