Question

at 25 degree celsius the vapor pressure of pure water is 27.76 mmHg and that of a dilute solution aqueous urea solution is 22.98 mmHg. estimate the molality of solution.

Answer 1

Question#47012 – Chemistry – Inorganic Chemistry

Question:

At 25 degree Celsius the vapor pressure of pure water is $27.76\ \mathrm{mmHg}$ and that of a dilute solution aqueous urea solution is $22.98\ \mathrm{mmHg}$. Estimate the molality of solution.

Answer:

Vapor pressure of solvent decreases with the presence of solute because of the reduction of evaporation surface. This phenomenon is illustrated with a picture below.

The decrease in vapor pressure is defined by Raoult’s law:

$$p_{\text{solvent}} = \chi_{\text{solvent}} \times p_{\text{solvent}}^{0} \tag{1}$$

where $p_{\text{solvent}}^{0}$ is a vapor pressure under pure solvent and $\chi$ is a molar fraction, defined as:

$$\chi_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}} \tag{2}$$

One can calculate the molar fraction of the given solution from equation (1):

$$\chi_{\text{solvent}} = \frac{p_{\text{solvent}}}{p_{\text{solvent}}^{0}} = \frac{22.98\ \mathrm{mmHg}}{27.76\ \mathrm{mmHg}} = 0.8278$$

Molality can be defined as the amount of moles of solute per one kilogram of solvent:

$$C_{m} = \frac{1000 \times n_{\text{solute}}}{m_{\text{solvent}}} \tag{3}$$

Equation (3) is for mass in grams.

We can show that the molar fraction of solute is related to molar fraction of solvent with the following equation:

$$\chi_{\text{ solute}} = \frac{n_{\text{ solute}}}{n_{\text{ solvent}} + n_{\text{ solute}}} = 1 - \chi_{\text{ solvent}} \tag{4}$$

Molality (3) is related with molar fraction of solute (4) according to the equation:

$$C_{m} = \frac{1000\chi_{\text{ solute }}}{(1 - \chi_{\text{ solute }})M_{\text{ solute }}} = \frac{1000(1 - \chi_{\text{ solvent }})}{\chi_{\text{ solvent }}M_{\text{ solvent }}} \tag{5}$$

In our case solvent is water $\mathrm{H}_2\mathrm{O}$ and solute is urea $\mathrm{CO(NH_2)_2}$. According to the equation (5):

$$C_{m}\left(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\right) = \frac{1000\left(1 - \chi_{\mathrm{H}_{2}\mathrm{O}}\right)}{\chi_{\mathrm{H}_{2}\mathrm{O}}M_{\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}}} = \frac{1000 \times (1 - 0.8278)}{0.8278 \times 18\ \mathrm{g/mol}} = 3.47\ \mathrm{mol/kg}$$