$^{-1}$) dissolved in 100 g of water at 25°C is 2 g. Calculate the following:

1. Molality of the solution

Molality is defined as the number of moles of solute per kilogram of solvent.

First, we need to calculate the number of moles of the solute:

n = m/M = 2 g / 50 g mol$^{-1}$ = 0.04 mol

Next, we need to calculate the mass of water in the solution:

mass of water = total mass - mass of solute = 100 g - 2 g = 98 g

Finally, we can calculate the molality:

molality = n / mass of solvent (in kg) = 0.04 mol / 0.098 kg = 0.408 mol kg$^{-1}$

Therefore, the molality of the solution is 0.408 mol kg$^{-1}$.

2. Mole fraction of the solute

The mole fraction of the solute is defined as the number of moles of the solute divided by the total number of moles in the solution.

First, we need to calculate the total number of moles in the solution:

n$_{total}$ = m$_{water}$/M$_{water}$ = 98 g / 18.015 g mol$^{-1}$ = 5.43 mol

Next, we can calculate the mole fraction of the solute:

x$_{solute}$ = n$_{solute}$ / n$_{total}$ = 0.04 mol / 5.43 mol = 0.00736

Therefore, the mole fraction of the solute is 0.00736.

3. Boiling point elevation of the solution

The boiling point elevation of a solution is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent. It is given by the equation:

ΔT$_b$ = K$_b$ x molality

where ΔT$_b$ is the boiling point elevation, K$_b$ is the molal boiling point elevation constant (for water, K$_b$ = 0.512°C kg mol$^{-1}$), and molality is the molality of the solution that we calculated in part 1.

ΔT$_b$ = K$_b$ x molality = 0.512°C kg mol$^{-1}$ x 0.408 mol kg$^{-1}$ = 0.209°C

Therefore, the boiling point of the solution is 0.209°C higher than the boiling point of pure water.

4. Freezing point depression of the solution

The freezing point depression of a solution is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution. It is given by the equation:

ΔT$_f$ = K$_f$ x molality

where ΔT$_f$ is the freezing point depression, K$_f$ is the molal freezing point depression constant (for water, K$_f$ = 1.86°C kg mol$^{-1}$), and molality is the molality of the solution that we calculated in part 1.

ΔT$_f$ = K$_f$ x molality = 1.86°C kg mol$^{-1}$ x