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

• Calculate the freezing point depression (\( \Delta T_f \)):

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

• Using the formula for freezing point depression:

\( \Delta T_f = K_f \cdot m = 2.5 = 1.86 \times \frac{n}{0.1} \)

• Solve for moles of methanol (\( n \)):

\( n = 0.1344 \, \text{moles} \)

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

\( w = 0.1344 \times 32 = 4.3 \, \text{g} \)

• Calculate the volume of methanol:

\( \text{Volume} = \frac{4.3}{0.792} = 5.43 \, \text{mL} = 543 \times 10^{-2} \, \text{mL} \)

Answer: \( x = 543 \)

Answer 2

The problem asks to find the value of 'x' to the nearest integer. A volume of \(x \times 10^{-2}\) mL of methanol is added to 100 mL of water, causing a depression in the freezing point, which is depicted in the given phase diagram.

Concept Used:

The problem is based on the colligative property of depression in freezing point. When a solute is added to a pure solvent, the freezing point of the solvent decreases. The depression in freezing point (\(\Delta T_f\)) is directly proportional to the molality (\(m\)) of the solution. The relationship is given by the formula:

\[ \Delta T_f = K_f \times m \]

where:

• \(\Delta T_f\) is the depression in freezing point, calculated as \(\Delta T_f = T_f^\circ - T_f\), where \(T_f^\circ\) is the freezing point of the pure solvent and \(T_f\) is the freezing point of the solution.
• \(K_f\) is the molal freezing point depression constant (cryoscopic constant) of the solvent.
• \(m\) is the molality of the solution, defined as:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \]

Step-by-Step Solution:

Step 1: Determine the depression in freezing point (\(\Delta T_f\)) from the given phase diagram.

From the diagram:

Freezing point of pure water, \(T_f^\circ = 273.15 \text{ K}\)

Freezing point of the methanolic solution, \(T_f = 270.65 \text{ K}\)

\[ \Delta T_f = T_f^\circ - T_f = 273.15 \text{ K} - 270.65 \text{ K} = 2.5 \text{ K} \]

Step 2: Calculate the molality (\(m\)) of the methanolic solution.

Using the formula for freezing point depression and the given value of \(K_f = 1.86 \text{ K kg mol}^{-1}\):

\[ m = \frac{\Delta T_f}{K_f} = \frac{2.5 \text{ K}}{1.86 \text{ K kg mol}^{-1}} \approx 1.344 \text{ mol kg}^{-1} \]

Step 3: Calculate the mass of the solvent (water) in kg.

Given volume of water = 100 mL and density of water = 1 g/cm³ (or 1 g/mL).

\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 100 \text{ mL} \times 1 \text{ g/mL} = 100 \text{ g} \]

Converting the mass to kilograms:

\[ \text{Mass of water} = \frac{100 \text{ g}}{1000 \text{ g/kg}} = 0.1 \text{ kg} \]

Step 4: Calculate the number of moles of solute (methanol) added.

Using the definition of molality:

\[ \text{Moles of methanol} = m \times \text{Mass of water (kg)} \] \[ \text{Moles of methanol} = 1.344 \text{ mol kg}^{-1} \times 0.1 \text{ kg} = 0.1344 \text{ mol} \]

Step 5: Calculate the mass of methanol corresponding to these moles.

Given molar mass of methanol (\(CH_3OH\)) = 32 g/mol.

\[ \text{Mass of methanol} = \text{Moles} \times \text{Molar mass} \] \[ \text{Mass of methanol} = 0.1344 \text{ mol} \times 32 \text{ g/mol} \approx 4.3008 \text{ g} \]

Step 6: Calculate the volume of methanol added.

Given density of methanol = 0.792 g/cm³ (or 0.792 g/mL).

\[ \text{Volume of methanol} = \frac{\text{Mass}}{\text{Density}} = \frac{4.3008 \text{ g}}{0.792 \text{ g/mL}} \approx 5.4303 \text{ mL} \]

Final Computation & Result:

Step 7: Equate the calculated volume to the given expression to find the value of x.

The problem states that the volume of methanol added is \(x \times 10^{-2}\) mL.

\[ x \times 10^{-2} \text{ mL} = 5.4303 \text{ mL} \] \[ x = \frac{5.4303}{10^{-2}} = 5.4303 \times 100 \] \[ x = 543.03 \]

The value of x to the nearest integer is 543.