Question

What mass of non-volatile solute (urea) needs to be dissolved in 100 gm of water in order to decrease the V.P. of water by 25%? What will be the molality of the solution?

Answer 1

Mass = 111.11 gm, Molality = 18.52 mol/kg

Answer 2

The relative lowering of vapor pressure of a solution containing a non-volatile solute is given by Raoult's Law:

$\frac{P_1^0 - P_1}{P_1^0} = x_2$

where $P_1^0$ is the vapor pressure of the pure solvent, $P_1$ is the vapor pressure of the solution, and $x_2$ is the mole fraction of the solute.

The problem states that the vapor pressure of water is decreased by 25%.

$\frac{P_1^0 - P_1}{P_1^0} = 0.25$

According to Raoult's Law, the mole fraction of the solute ($x_2$) is 0.25.

$x_2 = \frac{n_2}{n_1 + n_2} = 0.25$

where $n_1$ is the number of moles of the solvent (water) and $n_2$ is the number of moles of the solute (urea).

We are given the mass of water ($W_1$) = 100 gm. The molar mass of water ($M_1$) = 18 g/mol. The number of moles of water ($n_1$) is:

$n_1 = \frac{W_1}{M_1} = \frac{100 \text{ gm}}{18 \text{ g/mol}} = \frac{50}{9}$ moles.

Let the mass of urea be $W_2$. The molar mass of urea ($M_2$) = 60 g/mol. The number of moles of urea ($n_2$) is:

$n_2 = \frac{W_2}{M_2} = \frac{W_2}{60}$ moles.

Substitute the number of moles into the mole fraction equation:

$\frac{n_2}{n_1 + n_2} = 0.25$

$\frac{n_2}{50/9 + n_2} = 0.25$

$n_2 = 0.25 \left( \frac{50}{9} + n_2 \right)$

$n_2 = 0.25 \times \frac{50}{9} + 0.25 n_2$

$n_2 - 0.25 n_2 = 0.25 \times \frac{50}{9}$

$0.75 n_2 = \frac{1}{4} \times \frac{50}{9}$

$\frac{3}{4} n_2 = \frac{50}{36} = \frac{25}{18}$

$n_2 = \frac{25}{18} \times \frac{4}{3} = \frac{100}{54} = \frac{50}{27}$ moles.

Now, calculate the mass of urea ($W_2$) using the number of moles of urea:

$W_2 = n_2 \times M_2 = \frac{50}{27} \text{ moles} \times 60 \text{ g/mol} = \frac{50 \times 60}{27} = \frac{3000}{27} = \frac{1000}{9}$ gm.

$W_2 = 111.11$ gm (approximately).

Next, calculate the molality of the solution. Molality ($m$) is defined as the number of moles of solute per kilogram of solvent.

Molality ($m$) = $\frac{\text{Number of moles of solute (urea)}}{\text{Mass of solvent (water) in kg}}$

Number of moles of solute ($n_2$) = $\frac{50}{27}$ moles. Mass of solvent (water) = 100 gm = 0.1 kg.

Molality ($m$) = $\frac{50/27 \text{ moles}}{0.1 \text{ kg}} = \frac{50}{27 \times 0.1} = \frac{50}{2.7} = \frac{500}{27}$ mol/kg.

Molality ($m$) = 18.5185 mol/kg (approximately).

The mass of non-volatile solute (urea) needed is $\frac{1000}{9}$ gm or approximately 111.11 gm. The molality of the solution will be $\frac{500}{27}$ mol/kg or approximately 18.52 mol/kg.