What is the vapor pressure at 35°C of a solution made by dissolving 44.6 g of sucrose, C12H22O11, in 62.3 g of water? The vapor pressure of pure water at 35°C is 42.2 torr.

The correct answer is 40.861 torr.

Explanation:

Based on the given information, the weight of the solute (sucrose) given is 44.6 grams, the weight of the solvent, that is, water given is 62.3 grams, and 42.2 torr is the vapor pressure of pure water.

Now the molecular weight of sucrose is 342 grams per mole, and the molecular weight of water is 18 grams per mole.

The mole fraction of solute will be,

Mole fraction of solute = Weight of solute/Molecular weight of the solute/Weight of the solvent/molecular weight of the solvent + Weight of the solute/Molecular weight of the solute

= 44.6 grams/342 grams per mole/62.3 grams/18 grams per mole + 44.6 grams/342 grams per mole

= 0.1304/3.461 + 0.1304

= 0.0363

Now using relative lowering of vapor pressure,

Vapor pressure of pure water-Vapor pressure of solution/Vapor pressure of pure water = Mole fraction of solute

42.4 torr - PS/42.4 torr = 0.0363

Ps = 42.4 -1.5395

Ps = 40.861 torr

Hence, the vapor pressure of the solution is 40.861.

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