Question

ve the solution is ________ torr. The vapor pressure of pure water at 55 °C is 118.0 torr. The MW of lactose is 342.3 g/mol. A solution is prepared by adding 50.00 g of lactose (milk sugar) to 110.0 g of water at 55 °C. The partial pressure of water above the solution is ________ torr. The vapor pressure of pure water at 55 °C is 118.0 torr. The MW of lactose is 342.3 g/mol. 156.8 2.757 115.2 282.3 81.1

Answer 1

Answer:

$\boxed{\text{115.2 torr}}$

Explanation:

Let’s call water Component 1 and lactose Component 2.

According to Raoult’s Law,  

$p_{1} = \chi_{1}p_{1}^{\circ} \text{ and}\\p_{2} = \chi_{2}p_{2}^{\circ}$

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Data:

m₁ = 110.0 g;    p₁° = 118.0 torr

m₂ = 50.00 g; p₂° =    0    torr

1. Calculate the moles of each component

$n_{1} = \text{110.0 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{6.104 mol}\\\\n_{2} = \text{50.00 g} \times \dfrac{\text{1 mol}}{\text{342.3 g}} = \text{0.1461 mol}$

2. Calculate the mole fraction of each component

$\begin{array}{rcl}\chi_{2} & = & \dfrac{n_{2}}{n_{1} + n_{2}}\\\\&= & \dfrac{0.1461}{6.104 + 0.1461}\\\\& = &\dfrac{0.1461}{6.250}\\\\& = & \mathbf{0.023 37}\\\chi_{1}& = &1 - \chi_{2}\\& = &1 - 0.023 37\\& = & \mathbf{0.976 63}\\\end{array}$

3. Calculate the vapour pressure of the mixture

$p_{1} = 0.97663 \times \text{118.0 torr} = \text{ 115.2 torr}\\p_{2} = 0\\p_{\text{tot}} = p_{1} + p_{2} = \text{115.2 torr + 0} = \textbf{115.2 torr}\\\text{The partial pressure of water above the solution is \boxed{\textbf{115.2 torr}} }$