Question

How many grams of sucrose, C12H22O11, (MM= 342.3 g/mol), must be added to 631. g of water at 20°C to lower the vapor pressure by 0.963 mm Hg? The vapor pressure of water at 20°C is 17.5 mmHg

Answer 1

The fraction of vapor pressure lowering is 0.963 mmHg/17.5 mmHg = 0.05503. This should be equal to the mole fraction of the sucrose in the water. The number of moles of water is 631 g / 18.02 g/mol = 35.02 moles. If we have x grams of sucrose, then we have x/342.3 moles of sucrose, and the mole fraction of sucrose would be equal to 0.05503:
(x/342.3) / (x/342.3 + 35.02) = 0.05503
x = 698.0789 g sucrose

Answer 2

$\boxed{698.3\text{ g}}$ of sucrose must be added to 631 g of water to lower the vapor pressure by 0.963 mm Hg.

Further Explanation:

Colligative properties

These are the properties depending only on solute concentration, irrespective of their identities. Four colligative properties are mentioned below.

1. Relative lowering of vapor pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

Relative lowering of vapor pressure indicates the decrease in vapor pressure of solution when non-volatile solute is added to it. Due to this addition, vapor pressure decreases and this property is therefore named so.

The formula to calculate relative lowering of vapor pressure is as follows:

$\dfrac{p_1^0-p_1}{p_1^0}=x_2$                                                                                ...... (1)

Here,

$p_1^0$is the pressure of the pure water.

$p_1$is the pressure of the solution.

$x_2$ is the mole fraction of sucrose.

The formula to calculate the mole fraction of sucrose is as follows:

$x_2=\dfrac{n_2}{n_1+n_2}$                                                                          ....... (2)

Here,

$x_2$ is the mole fraction of sucrose.

$n_2$ is the number of moles of sucrose.

$n_1$ is the number of moles of water.

Incorporating equation (2) in equation (1), modified equation becomes;

$\dfrac{p_1^0-p_1}{p_1^0}=\dfrac{n_2}{n_2+n_1}$                                                   ...... (3)

The formula to calculate number of moles is as follows:

$\text{Number of moles}=\dfrac{\text{Mass}}{\text{Molar mass}}$                                       ...... (4)

Substitute 631 g for mass and 18.02 g/mol for molar mass in equation (4) to calculate number of moles of water.

$\begin{aligned}\text{Moles of water}\left(n_1\right)&=\dfrac{631\text{ g}}{18.02\text{ g/mol}}\\&=35.01\text{ mol}\end{aligned}$

Substitute 17.5 mm Hg for $p_1^0$, 0.963 mm Hg for $\left(p_1-p_1^o\right)$ and 35.01 mol for $n_1$in equation (3).

$\begin{aligned}\dfrac{0.963\text{ mm Hg}}{17.5\text{ mm Hg}}&=\dfrac{n_2}{35.01\text{ mol}+n_2}\\0.055&=\dfrac{n_2}{35.01\text{ mol}+n_2}\end{aligned}$

Solving for $n_2$ ,

$n_2=2.04\text{ mol}$

Rearrange equation (4) for mass.

$\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$                               ...... (5)

Substitute 2.04 mol for number of moles and 342.3 g/mol for molar mass in equation (5) to calculate mass of sucrose.

$\begin{aligned}\text{Mass of sucrose}&=\left(2.04 \text{ mol}\right)\left(342.3 \text{ g/mol}\right)\\&=698.3 \text{ g}\end{aligned}$

Learn more:

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Colligative properties

Keywords: colligative properties, relative lowering of vapor pressure, n1, n2, x2, p1, sucrose, water, 698.3 g, 2.04 mol, 342.3 g/mol, mass of sucrose, mass, molar mass.