Question

A solution is prepared by dissolving \(136.8\) g sucrose in \(1\) kg of water. Find the vapor pressure of solution at \(298\) K. [Molar mass of sucrose = \(342\) g \(\text{mol}^{-1}\), Vapour pressure of water = \(18\) mm Hg]

• A. \(17.75\ \text{mm Hg}\)
• B. \(17.79\ \text{mm Hg}\)
• C. \(17.95\ \text{mm Hg}\)
• D. \(17.87\ \text{mm Hg}\)

Hint:

Use Raoult's law with the mole fraction of water.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a non-volatile solute, \(P=P^{0}x_{\text{solvent}}\), or equivalently \(\dfrac{P^{0}-P}{P^{0}}=x_{\text{solute}}\).

Step 2: Find the Moles
\(n_{\text{sucrose}}=\dfrac{136.8}{342}=0.4\)
\(n_{\text{water}}=\dfrac{1000}{18}=55.56\)

Step 3: Mole Fraction
\(x_{\text{sucrose}}=\dfrac{0.4}{0.4+55.56}=0.00715\)

Step 4: Calculation
\(P=18\times(1-0.00715)=17.87\ \text{mm Hg}\), option (D).

Final Answer:
The vapour pressure of the solution is 17.87 mm Hg. \[ \boxed{\text{(D) }17.87\ \text{mm Hg}} \]