Question

Which of the following is the correct relation between osmotic pressure of 0.1M $NaCl$ solution and 0.1M $N{a_2}S{O_4}$ solution?

Answer 1

Hint: Osmotic pressure of salt solution depends on concentration of solution and also depends on the number of ions generated when one molecule of salt is dissolved with water. To make solution we dissolve salt in water and salts breakdowns into ions after dissolving with water and osmotic pressure is dependent on the concentration of ions in solution.

Complete step by step answer:
We know that Osmotic pressure of solution is given by
$\pi = i \times cRT$,
Here, Van't Hoff factor $i$ is $i = 1 + (n - 1)\alpha $ and $\alpha $is degree of dissociation.
For $NaCl$ and $N{a_2}S{O_4}$, $\alpha = 1$ because we know that both are ionic compounds and dissociate completely when dissolved in water.

For $NaCl$, in aqueous solution it gives one sodium and one chloride ion then, n=2.
$NaC{l_{(aq)}} \to N{a^ + }_{(aq)} + C{l^ - }_{(aq)}$
${\pi _{(NaCl)}} = 2 \times 0.1RT = 0.2RT$ -(1)

For $N{a_2}S{O_4}$, in aqueous solution it dissociates into two sodium ions and one sulphate, there are a total three ions then, n=3.
$N{a_2}S{O_4}_{(aq)} \to 2N{a^ + }_{(aq)} + S{O_4}{^{2 - }_{(aq)}}$ or n=2
${\pi _{(N{a_2}S{O_4})}} = 3 \times 0.1RT = 0.3RT$ -(2)

From equation (1) and (2), we come to know that
${\pi _{(NaCl)}} < {\pi _{(N{a_2}S{O_4})}}$

So, the correct answer is option B (osmotic pressure of $N{a_2}S{O_4}$ is more than $NaCl$ solution).


Note:
Here $\alpha = 1.$ because $N{a_2}S{O_4}$ and $NaCl$ are ionic solid and they dissociate completely into ions when we add them into water. Osmotic pressure can be defined as the minimum pressure applied to a solution to stop the flow of solvent molecules through a semipermeable membrane (osmosis).