Answer:
[tex]\boxed{\text{115.2 torr}}[/tex]
Explanation:
Let’s call water Component 1 and lactose Component 2.
According to Raoult’s Law,
[tex]p_{1} = \chi_{1}p_{1}^{\circ} \text{ and}\\p_{2} = \chi_{2}p_{2}^{\circ}[/tex]
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
[tex]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}[/tex]
2. Calculate the mole fraction of each component
[tex]\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}[/tex]
3. Calculate the vapour pressure of the mixture
[tex]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 $\boxed{\textbf{115.2 torr}}$}[/tex]
