Question

The osmotic pressure of $30\% $ (weight/volume) of anhydrous ${{Ca}}{{{Cl}}_2}$ solution at ${0^ \circ }{{C}}$, assuming $100\% $ dissociation is about:
A. $180{{atm}}$
B. $120{{atm}}$
C. $200{{atm}}$
D. $250{{atm}}$

Answer 1

Hint: Colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. There are four colligative properties. Osmotic pressure is one of the colligative properties.

Complete step-by-step answer:
Osmosis is the movement of water or other solvent through a plasma membrane from a region of low solute concentration to a region of high concentration. It does not require any energy to be applied.
The law of osmotic pressure is the combination of Boyle-van’t Hoff law and van’t Hoff-Charles’ law. Boyle-van’t Hoff law tells that the osmotic pressure of a solution at constant temperature is directly proportional to its concentration. Van’t Hoff-Charles’ law says that for a solution of fixed concentration the osmotic pressure of a solution is directly proportional to its absolute temperature.
The reaction showing the dissociation of ${{Ca}}{{{Cl}}_2}$ is given below:
${{Ca}}{{{Cl}}_2} \rightleftharpoons {{Ca^{2 + }} + 2{{Cl}}^ - }$
$1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$ At initial concentration
$1 - \alpha\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \alpha\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\alpha $ After dissociation
It is given that the mass by percentage composition of ${{Ca}}{{{Cl}}_2}$ is $30\% $.
i.e. $100{{mL}}$ of anhydrous ${{Ca}}{{{Cl}}_2}$ solution contains $30{{g}}$.
So mass of ${{Ca}}{{{Cl}}_2}$, ${{w = 30g}}$, volume, ${{V = 100mL}}$, temperature, ${{T = }}{{{0}}^ \circ }{{C = 273K}}$
Concentration of the solution in molar concentration, ${{C}}$ can be calculated by dividing the number of moles (mass divided by molar mass) of ${{Ca}}{{{Cl}}_2}$ by volume.
${{C = }}\dfrac{{{{w}} \times {{1000}}}}{{{{W}} \times {{V}}}}$, where ${{W}}$ is the molar mass of ${{Ca}}{{{Cl}}_2}$ $\left( {111{{g}}.{{mol}}^{ - 1}} \right)$
${{C = }}\dfrac{{30{{g}} \times 1000}}{{111{{g}}.{{mol}}^{ - 1}} \times 100{{mL}}} = 2.7{{M}}$
From these values, we can calculate the osmotic pressure.
Osmotic pressure, $\pi = {{iCRT}}$, where ${{R}}$ is the gas constant $\left( {0.0821{{L}}.{{atm}}.{{mol}}^{ - 1}}.{{{K}}^{ - 1}} \right)$ and ${{i}}$ is the total number of moles after dissociated $\left( {{{i = 3}}} \right)$
$\pi = 3 \times 2.7{{M}} \times 0.0821{{L}}.{{atm}}.{{mol}}^{ - 1}.{{{K}}^{ - 1}} \times 273{{K = 181}}{{.53atm}}$
Hence the closest number from the given options is $180{{atm}}$.

So the correct option is A.

Note: Van’t Hoff factor is a measure of the effect of solute upon all the colligative properties. Osmotic pressure in a diluted solution is equal to the pressure exerted by the solute if it is a gas occupying the same volume.