Question

1.95 g of non-volatile and non-electrolyte solute dissolved in 100 g of benzene lowered the freezing point of it by 0.64 K. The molar mass of the solute (in g mol\(^{-1}\)) (\(K_f(\text{C}_6\text{H}_6) = 5.12 \, \text{K kg mol}^{-1}\)) is

• A. 240
• B. 156
• C. 165
• D. 265

Hint:

- Depression in freezing point: \( \Delta T_f = K_f \cdot m \). - Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{w_2/M_2}{w_1/1000} = \frac{w_2 \times 1000}{M_2 \times w_1} \) (if \(w_1\) is in grams). - So, \( M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} \). - \(K_f\) is the molal freezing point depression constant or cryoscopic constant. - Ensure units are consistent. \( \Delta T_f \) is in K or \(^\circ\)C (as it's a difference).

Answer 1

The Correct Option is B

The depression in freezing point \( \Delta T_f \) is given by the formula: \[ \Delta T_f = K_f \cdot m \] where \( K_f \) is the molal freezing point depression constant (cryoscopic constant) of the solvent, and \(m\) is the molality of the solution.
Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \).
Let \( w_2 \) be the mass of solute, \( M_2 \) be the molar mass of solute, and \( w_1 \) be the mass of solvent.
Then, moles of solute \( = \frac{w_2}{M_2} \).
Molality \( m = \frac{w_2/M_2}{w_1 (\text{in kg})} = \frac{w_2 \times 1000}{M_2 \times w_1 (\text{in g})} \).
So, \( \Delta T_f = K_f \cdot \frac{w_2 \times 1000}{M_2 \times w_1} \).
We need to find \( M_2 \).
Rearranging the formula: \[ M_2 = K_f \cdot \frac{w_2 \times 1000}{\Delta T_f \times w_1} \] Given values: Mass of solute \( w_2 = 1.
95 \) g.
Mass of solvent (benzene) \( w_1 = 100 \) g.
Depression in freezing point \( \Delta T_f = 0.
64 \) K.
Cryoscopic constant for benzene \( K_f = 5.
12 \, \text{K kg mol}^{-1} \).
Substitute the values: \[ M_2 = (5.
12 \, \text{K kg mol}^{-1}) \cdot \frac{(1.
95 \, \text{g}) \times 1000}{(0.
64 \, \text{K}) \times (100 \, \text{g})} \] Units check: K cancels.
kg from \(K_f\) cancels with kg from converting \(w_1\) to kg (implicit in the \(1000/w_1\) factor).
Result will be in g/mol.
\[ M_2 = 5.
12 \times \frac{1.
95 \times 10}{0.
64} \] \[ M_2 = \frac{5.
12 \times 19.
5}{0.
64} \] Notice that \( 5.
12 / 0.
64 = 512 / 64 \).
\( 64 \times 8 = 512 \).
So, \( \frac{5.
12}{0.
64} = 8 \).
\[ M_2 = 8 \times 19.
5 \] \[ M_2 = 8 \times (20 - 0.
5) = 160 - 8 \times 0.
5 = 160 - 4 = 156 \] The molar mass of the solute is \( 156 \, \text{g mol}^{-1} \).
This matches option (2).