A set of solutions is prepared using 180 g of water as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order :
[Given, molar mass of A = 100 g mol\(^{-1}\); B = 200 g mol\(^{-1}\); C = 10,000 g mol\(^{-1}\)]
[Given, molar mass of A = 100 g mol\(^{-1}\); B = 200 g mol\(^{-1}\); C = 10,000 g mol\(^{-1}\)]
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- A > C > B
- C > B > A
- A > B > C
- B > C > A
The Correct Option is C
Solution and Explanation
Moles of water \(= \frac{180}{18} = 10\).
Moles of A \(= \frac{10}{100} = 0.1\); B \(= \frac{10}{200} = 0.05\); C \(= \frac{10}{10000} = 0.001\).
\(x_A = \frac{0.1}{10.1} \approx 0.0099\); \(x_B = \frac{0.05}{10.05} \approx 0.00498\); \(x_C = \frac{0.001}{10.001} \approx 0.0001\).
So the order is A > B > C.
Answer: (3) A > B > C
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