Question

Calculate the vapor pressure of water above a solution prepared by adding 24.5 g of lactose (C12H22O11) to 200.0 g of water at 338 K. (Vapor-pressure of water at 338 K 187.5 torr.)

Answer 1

Answer:

Vapor pressure of solution → 186.3  Torr

Explanation:

To solve this problem we must apply the colligative property of vapor pressure.

ΔP =  P° . Xm

Where ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

P° is vapor pressure of pure solvent → 187.5 Torr

Xm is the mole fraction of solute (moles of solute / total moles)

Let's determine the mole fraction,

Moles of solute = Mass of solute / Molar mass

24.5 g / 342 g/mol = 0.0716 moles

Moles of solvent = Moles of solvent /Molar mass

200 g / 18g/mol = 11.1 moles

Total moles = 11.1 moles + 0.0716 moles → 11.1716 moles

Mole fraction of solute = 0.0716 mol / 11.1716 mol = 6.40×10⁻³

Let's apply the formula

ΔP =  P° . Xm

Vapor P of pure solvent - Vapor P of solution = P° . 6.40×10⁻³

Vapor P of pure solvent  - 187.5 Torr . 6.40×10⁻³ = Vapor P of solution

187.5 Torr - 187.5 Torr . 6.40×10⁻³ = Vapor P of solution

Vapor pressure of solution → 186.3  Torr

Answer 2

The vapor pressure has been the pressure exerted by the molecules in the solution in vapor phase. The vapor pressure of water in solution is 186.3 torr

What are colligative properties?

The colligative properties of the solution are dependent on the solute particles in the solution. The colligative properties include boiling point, freezing point, vapor pressure and osmotic pressure.

The change in the vapor pressure of the solution with the addition of solute ($\Delta P$) is given as :

$\Delta P=P^\circ\;\times\;x_m$

Where, the vapor pressure of the pure solvent, $P^\circ=187.5\;\rm torr$

The mole fraction of the solute ($x_m$) is given as:

$x_m=\rm \dfrac{Moles\;solute}{Total\;moles}$

The moles of lactose in 24.5 grams of sample is:

$\rm Moles\;solute=\dfrac{mass}{molar\;mass}\\\\ Moles\;lactose=\dfrac{24.5}{342\;g/mol} \\\\Moles\;Lactose=0.0706\;moles$

The moles of solvent water is given as:

$\rm Moles\;solvent\;(water)=\dfrac{200}{18\;g/mol} \\Moles\;solvent=11.1\;mol$

The total moles of the solution is given as:

$\rm Total \;moles=solute+solvent\\Total\;moles=0.0706+11.1\;mol\\Total \;moles=11.1706\;mol$

The moles fraction of solute is given as:

$x_m=\dfrac{0.0706}{11.1706}\\\\ x_m=6.40\;\times\;10^{-3}$

The vapor pressure change in the solution is calculated as:

$\Delta P=187.5\;\times\;6.40\;\times\;10^-^3\\\Delta P=1.2\;\rm torr$

The vapor pressure of the solution is given as:

$\Delta P=\rm Solution\;vapor\;pressure-solvent\;vapor\;pressure\\1.2=Solution\;vapor\;pressure-187.5\;torr\\Solution\;vapor\;pressure=186.3\;torr$

The vapor pressure of the solution is 186.3 torr.

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