Question

Let \( x \) grams of benzoic acid (molar mass = 122 g/mol\(^{-1}\)) be dissolved in 50 g of benzene. Its freezing point was found to be 277.82 K. What is the value of \( x \)? (Given: \( K_f \) of benzene = 5.1 K kg mol\(^{-1}\), freezing point of benzene = 278.45 K, and van't Hoff's factor of benzoic acid = 0.5)

• A. \(0.5 \)
• B. \(1.5 \)
• C. \(0.75 \)
• D. \(1.0 \)

Hint:

Remember, the freezing point depression depends directly on the molality and the nature of the solute. Benzoic acid, with a van't Hoff factor of 0.5, shows the effect of dissociation or association in solution.

Answer 1

The Correct Option is B

Given values are:

• Freezing point of pure benzene, \(T_f^0 = 278.45 \, \text{K}\)
• Freezing point of solution, \(T_f = 277.82 \, \text{K}\)
• \(K_f\) for benzene = \(5.1 \, \text{K kg mol}^{-1}\)
• Van't Hoff factor, \(i = 0.5\)

Step 1: Calculate the freezing point depression (\(\Delta T_f\)): \[ \Delta T_f = T_f^0 - T_f \] \[ = 278.45 \, \text{K} - 277.82 \, \text{K} \] \[ = 0.63 \, \text{K} \] Step 2: Use the freezing point depression formula to find the molality (\(m\)): \[ m = \frac{\Delta T_f}{i \times K_f} \] \[ = \frac{0.63 \, \text{K}}{0.5 \times 5.1 \, \text{K kg mol}^{-1}} \] \[ = \frac{0.63}{2.55} \, \text{mol/kg} \] Step 3: Define molality in terms of solute mass: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \] Let \(x\) be the mass of benzoic acid in grams. Moles of benzoic acid = \(\frac{x}{122}\) (Molar mass of benzoic acid = 122 g/mol). Mass of solvent (benzene) = 50 g = 0.05 kg. Substituting these values into the molality equation: \[ m = \frac{x/122}{0.05} = \frac{x}{122 \times 0.05} \] Step 4: Solve for \(x\): \[ \frac{x}{122 \times 0.05} = \frac{0.63}{2.55} \] \[ x = \frac{0.63}{2.55} \times 122 \times 0.05 \] \[ x = \frac{0.63 \times 122 \times 0.05}{2.55} \] \[ x \approx 1.50666... \] Rounding to one decimal place, \(x \approx 1.5 \, \text{g}\). Correct Answer: (2) 1.5