Question

At T K, vapour pressures of H\(_2\)O and CH\(_3\)OH are 36 and 120 torr respectively. 72 g of H\(_2\)O and x g of CH\(_3\)OH when mixed, the resulting solution has a total vapour pressure of 64 torr. Find x.

• A. 32
• B. 56
• C. 64
• D. 72

Hint:

Always convert mass → moles first before applying Raoult’s law for mixed solutions.

Answer 1

The Correct Option is C

Step 1: Apply Raoult’s law.
For an ideal solution: \[ P_{total} = P_1 x_1 + P_2 x_2 \] where \(x_1, x_2\) are mole fractions of components.

Step 2: Convert given masses into moles.
For water: \[ n_{H_2O} = \frac{72}{18} = 4 \, mol \] For methanol: \[ n_{CH_3OH} = \frac{x}{32} \]

Step 3: Express mole fractions.
Total moles: \[ n = 4 + \frac{x}{32} \] \[ x_{H_2O} = \frac{4}{n}, \quad x_{MeOH} = \frac{x/32}{n} \]

Step 4: Apply vapour pressure equation.
\[ 64 = 36 \cdot \frac{4}{n} + 120 \cdot \frac{x/32}{n} \] Multiply both sides by \(n\): \[ 64n = 144 + \frac{120x}{32} \]

Step 5: Substitute \(n\) and solve.
\[ 64\left(4 + \frac{x}{32}\right) = 144 + \frac{120x}{32} \] Expanding: \[ 256 + 2x = 144 + \frac{120x}{32} \] Simplify: \[ 256 + 2x = 144 + 3.75x \] \[ 112 = 1.75x \] \[ x = 64 \]

Step 6: Final conclusion.
Thus, mass of CH\(_3\)OH is 64 g.
Final Answer: \[ \boxed{64} \]