Answer to Question #134300 in Chemistry for Patricia

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. Compute for its boiling point at 20 degree Celsius?

Answer 1

The osmotic pressure $\pi$ and the boiling point elevation $\Delta T_b$ 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.

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 boiling point elevation is:

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

Answer: the boiling point elevation is 0.016 °C.