# How do you calculate the vapor pressure of water above a solution prepared by adding 22.5 g of lactose [C_12H_22O_11 ]tp 200 grams of water at 338 K?

Jul 8, 2017

We need (i) the mole fractions of water in the solution; and (ii) the vapour pressure of pure water at $338 \cdot K$.....we finally get a small reduction of solution vapour pressure.

#### Explanation:

And (ii) we need the vapour pressure of pure water at this temperature. This site quotes the vapour pressure of PURE water at $338 \cdot K$ (i.e. $65$ ""^@C) as approx. $55 \cdot m m \cdot H g$. You should have quoted the EXACT value in the question.

Now the vapour pressure of the solution will be proportional to the mole fraction of each component.......

${\chi}_{{H}_{2} O} = \frac{\frac{200 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}}{\frac{200 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} + \frac{22.5 \cdot g}{342.3 \cdot g \cdot m o {l}^{-} 1}} = 0.994$

${\chi}_{{C}_{12} {H}_{22} {O}_{11}} = \frac{\frac{22.5 \cdot g}{342.3 \cdot g \cdot m o {l}^{-} 1}}{\frac{200 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} + \frac{22.5 \cdot g}{342.3 \cdot g \cdot m o {l}^{-} 1}} = 0.006$

And as is required for a binary solution, the sum of the mole fractions is equal to unity.......

${\chi}_{{H}_{2} O} + {\chi}_{{C}_{12} {H}_{22} {O}_{11}} = 1$

But $\text{lactose}$ is considered to be involatile and DOES not contribute to the vapour pressure. Had the solute been ethanol, or methanol, then this would have made some contribution to the overall vapour pressure.

And thus at $65$ ""^@C, the pure solvent will have a vapour pressure of $55 \cdot m m \cdot H g$. The solution will undergo vapour pressure diminution of $0.994 \times 55 \cdot m m \cdot H g = 54.6 \cdot m m \cdot H g$. And thus the vapour pressure is ONLY slightly reduced........