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Relative Lowering of Vapour Pressure

The vapor pre...

Question

The vapor pressure of water at $20 C$ is $17.5 m m H g$ . What is the vapor pressure of water over a solution prepared from $200 g$ of sucrose $(C_{12} H_{22} O_{11})$ and $112.3 g$ of water?

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Solution

We use Raoult's Law, and calculate a solution vapor pressure of $16 m m o f H g$ .

Explanation:
Raoult's Law tells us that in an ideal solution, the vapor pressure of a component is proportional to the mole fraction of that component in the solution.
Mole fraction of sucrose = Moles of sucrose / Moles of sucrose + moles of water.
$X_{s u c r o s e} = \frac{\frac{200 g}{342.13 g \cdot m o l^{- 1}}}{\frac{200 g}{342.13 g \cdot m o l^{- 1}} + \frac{112.3 g}{18.01 g \cdot m o l^{- 1}}} = 0.0857$
Mole fraction of water:
$X_{w a t e r} = \frac{\frac{112.3 g}{18.01 g \cdot m o l^{- 1}}}{\frac{200 g}{342.13 g \cdot m o l^{- 1}} + \frac{112.3 g}{18.01 g \cdot m o l^{- 1}}} = 0.914$
Note that by definition: $X_{s u c r o s e} + X_{w a t e r} = 1$

Because sucrose is involatile, the vapor pressure of the solution is proportional to the mole fraction of water:
$P_{s o l u t i o n} = X_{w a t e r} \times 17.5 m m o f H g$ $= 0.914 \times 17.5 m m o f H g = 16 m m o f H g$

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