Question:
A solution is prepared by dissolving \(68\) g sucrose in \(1\) kg water. Calculate the vapour pressure of solution at \(298\) K. [vapour pressure of water at \(298\) K = \(18\) mm of Hg & mole fraction of solvent = \(0.9964\)]
A solution is prepared by dissolving \(68\) g sucrose in \(1\) kg water. Calculate the vapour pressure of solution at \(298\) K. [vapour pressure of water at \(298\) K = \(18\) mm of Hg & mole fraction of solvent = \(0.9964\)]
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Raoult's law: \(p = x_{\text{solvent}}\,p^0\).
Updated On: Oct 1, 2026
- \(17.93\) mm Hg
- \(17.80\) mm Hg
- \(17.76\) mm Hg
- \(17.68\) mm Hg
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For a non-volatile solute like sucrose, only the solvent contributes to vapour pressure. Raoult's law gives \(p = x_1 p_1^0\).
Step 2: Calculate:
\[ p = 0.9964\times18 = 17.935\ \text{mm Hg} \approx 17.93\ \text{mm Hg} \]
Step 3: Check:
The drop is only about \(0.065\) mm Hg, because \(0.2\) mol of sucrose sits in \(55.5\) mol of water. The other options show a larger drop than this.
Final Answer:
The vapour pressure is \(17.93\) mm Hg, option (A). \[ \boxed{17.93\ \text{mm Hg}} \]
For a non-volatile solute like sucrose, only the solvent contributes to vapour pressure. Raoult's law gives \(p = x_1 p_1^0\).
Step 2: Calculate:
\[ p = 0.9964\times18 = 17.935\ \text{mm Hg} \approx 17.93\ \text{mm Hg} \]
Step 3: Check:
The drop is only about \(0.065\) mm Hg, because \(0.2\) mol of sucrose sits in \(55.5\) mol of water. The other options show a larger drop than this.
Final Answer:
The vapour pressure is \(17.93\) mm Hg, option (A). \[ \boxed{17.93\ \text{mm Hg}} \]
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