Question

At 300 K, 0.06 kg of an organic solute is dissolved in 1 kg of water. The vapour pressure of the solution is 3.768 kPa. If the vapour pressure of pure water at that temperature is 3.78 kPa, what is the molar mass of the solute (in g mol\(^{-1}\))?

• A. 180
• B. 120
• C. 340
• D. 260
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Hint:

Relative lowering of vapour pressure is a colligative property dependent on solute concentration, not identity.

Answer 1

The Correct Option is C

Step 1: Using the Relative Lowering of Vapour Pressure Formula
\[ \frac{P^0 - P}{P^0} = \frac{n_B}{n_A} \] where, \( P^0 \) = vapour pressure of pure water = 3.78 kPa
\( P \) = vapour pressure of solution = 3.768 kPa
\( n_B \) = moles of solute
\( n_A \) = moles of solvent
Step 2: Calculating Moles of Water
\[ n_A = \frac{1000}{18} = 55.56 \text{ moles} \] Step 3: Calculating Molar Mass of Solute
\[ \frac{3.78 - 3.768}{3.78} = \frac{\frac{0.06}{M}}{55.56} \] Solving for \( M \), \[ M = 340 \text{ g/mol} \] Thus, the molar mass of the solute is 340 g/mol.
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