Question

An aqueous solution of Nitric acid has an osmotic pressure of 0.80 atm at 27 degree Celsius. The density of solution is 1.03g/mol. What is the freezing point of the solution at 20 degree Celsius?

Answer 1

The osmotic pressure $\pi$ and the freezing point depression $\Delta T_f$ are the colligative properties of the solutions. The freezing point of the solution can be calculated from:

$\Delta T_f = -iK_fm$,

where $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.

On the other hand, the osmotic pressure of the solution can be calculated from:

$\pi = cRTi$ ,

where $c$ is the molarity, $R$ is the gas constant (8.314 J K^-1 mol^-1) and $T$ is the temperature in kelvin (27+273.15 = 300.15 K).

The nitric acid is a strong electrolyte, therefore we can assume that it dissociates completely with the formation of 2 ions:

HNO₃ $\rightarrow$ H⁺ + NO₃^-.

Thus, its van't Hoff factor $i$ equals 2.

Now, one can calculate the concentration of the solution of the nitric acid from the osmotic pressure given:

$c = \frac{\pi}{RTi} = \frac{0.8\cdot101325\text{ J/m}^3}{8.314\text{ J/Kmol}\cdot300.15\text{ K}\cdot2} = 16.24 \text{ mol/m}^3$ .

The molarity $c$ and the molality $m$ are linked through the density of the solution and the molar mass of the solute ($M$ for HNO₃ is 63.01 g/mol):

$m = \left[\frac{d_{sln}}{c} - M_s\right]^{-1} = 0.0158\text { mol/kg}$ .

Finally, the freezing point depression is:

$\Delta T_f = -2\cdot1.86\text{ (°C kg/mol) }\cdot0.0158\text{ (mol/kg)} =- 0.059$ °C.

Answer: the freezing point depression is -0.059 °C.