Chemistry · General chemistry II · Worked example

Vapor pressure of a glucose solution

$50.0 g$ of glucose, $C_{6} H_{12} O_{6}$ , is dissolved in $250.0 g$ of water at $25 ° C$ , where pure water’s vapor pressure is $23.8 torr$ . Find the solution’s vapor pressure and the lowering.

P_{solution} = χ_{water} P_{water}^{\circ}

Moles of each component

The first example used molality. Raoult’s law needs a mole fraction, so count both components in moles. Glucose is $180.16 g / mol$ and water is $18.02 g / mol$ . Glucose is nonvolatile and does not ionize.

n_{glucose} = \frac{50.0}{180.16} = 0.2775 mol, n_{water} = \frac{250.0}{18.02} = 13.87 mol

\begin{array}{c} n_{glucose} = \frac{50.0}{180.16} = 0.2775 mol \\ n_{water} = \frac{250.0}{18.02} = 13.87 mol \end{array}

Mole fraction of the solvent

Mole fraction counts particles, so use moles, not grams. Glucose is $16.7 %$ of the mass but under $2 %$ of the molecules.

χ_{water} = \frac{13.87}{13.87 + 0.2775} = 0.9804

Apply Raoult’s law

Multiply by the pure-water vapor pressure at the same temperature. The lowering is the difference, or equivalently $χ_{glucose} \times P °$ $= (0.0196) (23.8)$ $= 0.467 torr$ .

P = (0.9804) (23.8 torr) = 23.3 torr, Δ P = 0.47 torr

\begin{array}{c} P = (0.9804) (23.8 torr) \\ = 23.3 torr \\ Δ P = 0.47 torr \end{array}

Result

The vapor pressure is $23.3 torr$ , lowered by about $0.47 torr$ .

Your turn

1. One like the example

$30.0 g$ of sucrose, $C_{12} H_{22} O_{11}$ $(342.30 g / mol)$ , is dissolved in $100.0 g$ of water at $25 ° C$ , where $P ° = 23.8 torr$ . Find the solution’s vapor pressure.

Show the answer and explanation

$23.4 torr$ .

$30.0 g \div 342.30 g / mol$ $= 0.08764 mol$ of sucrose and $100.0 g \div 18.02 g / mol$ $= 5.549 mol$ of water. $χ (water)$ $= 5.549 \div 5.637$ $= 0.9845$ , and $P = 0.9845 \times 23.8 torr$ $= 23.4 torr$ .

\begin{array}{l} 30.0 g C_{12} H_{22} O_{11} \times \frac{1 mol C_{12} H_{22} O_{11}}{342.30 g C_{12} H_{22} O_{11}} = 0.08764 mol C_{12} H_{22} O_{11} \\ 100.0 g H_{2} O \times \frac{1 mol H_{2} O}{18.02 g H_{2} O} = 5.549 mol H_{2} O \\ \frac{5.549 mol}{5.549 mol + 0.08764 mol} = 0.9845 \\ (0.9845) (23.8 torr) = 23.4 torr \end{array}

\begin{array}{c} 30.0 g C_{12} H_{22} O_{11} \\ \times \frac{1 mol}{342.30 g} \\ = 0.08764 mol C_{12} H_{22} O_{11} \\ 100.0 g H_{2} O \times \frac{1 mol H_{2} O}{18.02 g H_{2} O} \\ = 5.549 mol H_{2} O \\ \frac{5.549 mol}{5.549 mol + 0.08764 mol} \\ = 0.9845 \\ (0.9845) (23.8 torr) \\ = 23.4 torr \end{array}

2. A variation

Would $0.2775 mol$ of $NaCl$ in the same $250.0 g$ of water lower the vapor pressure more or less than the glucose, and by about how much?

Show the answer and explanation

More, by about $0.916 torr$ , nearly twice as much.

Ideally $NaCl$ gives $2 \times 0.2775 = 0.555 mol$ of ions. $χ (solute particles)$ $= 0.555 \div (13.87 + 0.555)$ $= 0.0385$ , so $Δ P = 0.0385 \times 23.8 torr$ $= 0.916 torr$ . The glucose lowered it by $0.467 torr$ .

\begin{array}{l} (2) (0.2775 mol) = 0.555 mol \\ \frac{0.555 mol}{13.87 mol + 0.555 mol} = 0.0385 \\ (0.0385) (23.8 torr) = 0.916 torr \end{array}

\begin{array}{c} (2) (0.2775 mol) = 0.555 mol \\ \frac{0.555 mol}{(13.87 + 0.555) mol} \\ = 0.0385 \\ (0.0385) (23.8 torr) \\ = 0.916 torr \end{array}

3. Somewhere new

$10.0 g$ of glycerol, $C_{3} H_{8} O_{3}$ , a nonvolatile solute, is dissolved in $100.0 g$ of ethanol, $C_{2} H_{5} OH$ . Take pure ethanol’s vapor pressure as $100.0 torr$ at the temperature of the problem. Find the solution’s vapor pressure.

Show the answer and explanation

$95.2 torr$ .

Raoult’s law works for any solvent. $10.0 g \div 92.09 g / mol$ $= 0.1086 mol$ of glycerol and $100.0 g \div 46.07 g / mol$ $= 2.171 mol$ of ethanol. $χ (ethanol)$ $= 2.171 \div 2.280$ $= 0.9524$ , so $P = 0.9524 \times 100.0 torr$ $= 95.2 torr$ .

\begin{array}{l} 10.0 g C_{3} H_{8} O_{3} \times \frac{1 mol C_{3} H_{8} O_{3}}{92.09 g C_{3} H_{8} O_{3}} = 0.1086 mol C_{3} H_{8} O_{3} \\ 100.0 g C_{2} H_{5} OH \times \frac{1 mol C_{2} H_{5} OH}{46.07 g C_{2} H_{5} OH} = 2.171 mol C_{2} H_{5} OH \\ \frac{2.171 mol}{2.171 mol + 0.1086 mol} = 0.9524 \\ (0.9524) (100.0 torr) = 95.2 torr \end{array}

\begin{array}{c} 10.0 g C_{3} H_{8} O_{3} \\ \times \frac{1 mol}{92.09 g} \\ = 0.1086 mol C_{3} H_{8} O_{3} \\ 100.0 g C_{2} H_{5} OH \\ \times \frac{1 mol}{46.07 g} \\ = 2.171 mol C_{2} H_{5} OH \\ \frac{2.171 mol}{(2.171 + 0.1086) mol} \\ = 0.9524 \\ (0.9524) (100.0 torr) \\ = 95.2 torr \end{array}

Keep exploring

Open the solution in the Colligative studio’s Raoult mode, which gives $23.33 torr$ . Double the glucose to $0.555 mol$ and the lowering nearly doubles, to $0.916 torr$ .

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 10 key terms Check it in Colligative properties Check the mole fraction in a Chemistry box Open worked example on a board Chemistry formulas: solutions

Your existing work stays on this device. Examples open as editable copies.

Moles of each component

Mole fraction of the solvent

Apply Raoult’s law

Result

Your turn

1. One like the example

23.4 torr23.4 torr.

2. A variation

More, by about 0.916 torr0.916 torr, nearly twice as much.

3. Somewhere new

95.2 torr95.2 torr.

Keep exploring

Study it, then try it

$23.4 torr$ .

More, by about $0.916 torr$ , nearly twice as much.

$95.2 torr$ .