Question

Solute A is a ternary electrolyte and solute B is a none-electrolyte. If $0.1\mathrm{M}$solution of solute B produces an osmotic pressure of $2\mathrm{P}$, then $0.05\mathrm{M}$ solution of A at the same temperature will produce an osmotic pressure equal to?

Answer 1

Step 1: Analyzing the given data

The concentration of solute B$=0.1\mathrm{M}$

The osmotic pressure of solute B$=2\mathrm{P}$

The concentration of solute A$=0.05\mathrm{M}$

Step 2: Calculating osmotic pressure of solute A

Ternary electrolyte means which give three ions on dissociation, therefore their vant Hoff factor is three.

Now, Osmotic pressure ($\mathrm{\pi }$)$=\mathrm{CRT}\times \mathrm{i}$

For, Solute A $\mathrm{i}=3$ and $\mathrm{C}=0.05\mathrm{M}$

For Solute B$\mathrm{i}=1,\mathrm{C}=0.1,\mathrm{\pi }=2\mathrm{P}$

Given: Temperature is same

Therefore, equating both the osmotic pressures we get,

$$\begin{gathered} \frac{{\mathrm{\pi }}_1}{{\mathrm{C}}_1{\mathrm{i}}_1}=\frac{{\mathrm{\pi }}_2}{{\mathrm{C}}_2{\mathrm{i}}_2} \\ \mathrm{where},{\mathrm{\pi }}_1=\mathrm{osmotic}\mathrm{pressure}\mathrm{of}\mathrm{A},{\mathrm{\pi }}_2=\mathrm{osmotic}\mathrm{pressure}\mathrm{of}\mathrm{B} \\ {\mathrm{C}}_1=\mathrm{concentration}\mathrm{of}\mathrm{A},{\mathrm{C}}_2=\mathrm{concentration}\mathrm{of}\mathrm{B} \\ {\mathrm{i}}_1=\mathrm{vant}\mathrm{Hoff}\mathrm{factor}\mathrm{of}\mathrm{A},{\mathrm{i}}_2=\mathrm{vant}\mathrm{Hoff}\mathrm{factor}\mathrm{of}\mathrm{B} \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{\pi }}_1}{0.05\times 3}=\frac{2\mathrm{P}}{0.1} \\ {\mathrm{\pi }}_1=3\mathrm{P} \end{gathered}$$

Therefore, $0.05\mathrm{M}$ the solution of A at the same temperature will produce an osmotic pressure$=3\mathrm{P}$