Question

An aqueous solution of Nitric acid has an osmotic pressure of 0.80 atm at 27 deg-Celsius. The density of the solution is 1.03 g/mL.
a. What is the freezing point of the solution?
b. What is the boiling point of the solution?
c. What is the vapor pressure?

Answer 1

The osmotic pressure $\pi$, the boiling point elevation $\Delta T_b$ , the freezing point depression $\Delta T_f$ and relative lowering of vapour pressure $\Delta p$ are  the colligative properties of the solutions. The boiling point of the solution can be calculated from:

$\Delta T_b = iK_bm$,

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

The freezing point of the solution can be calculated from:

$\Delta T_f = -iK_fm$,

where $K_f$ is the cryoscopic constant.

The relative lowering of the vapour pressure $\Delta p$ can be calculated from the Raoult's law:

$\Delta p = p_wx_{HNO_3},$

where $p_w$ is the vapour pressure of pure water at a given temperature and $x_{HNO_3}$is the mole fraction of the nitric acid in 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):

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

Therefore, the boiling point elevation is:

$\Delta T_b = 2\cdot0.512\text{ (°C kg/mol) }\cdot0.0158\text{ (mol/kg)} = 0.016$ °C.

The freezing point depression is:

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

The mole fraction of the nitric acid is related to the molality of the solution through the molar mass of water (remember to take into account the dissociation of the nitric acid):

$x_{HNO_3} = \left[1+\frac{1}{2mM_w}\right]^{-1} = 0.00057$ .

Finally, the relative lowering of the vapour pressure is:

$\Delta p = 0.0231\text{ atm}\cdot0.00057 = 1.3\cdot10^{-5}\text{ atm}$

Answer: the boiling point elevation is 0.016 °C, the freezing point depression is -0.059 °C, the relative lowering of the vapour pressure at 20°C is 1.3x10^-5 atm.