Question

An aqueous solution of a substance molar mass $240$ has osmotic pressure $0.2atm$ at $300K$. The density of solution in $g.d{m^{ - 3}}$ is: ($R = 0.08\;litre\;atm\;{K^{ - 1}}mo{l^{ - 1}}$)

Answer 1

Hint: Density is the ratio between given mass and the volume of the solution. Thus, we can use this relation in the formula for osmotic pressure to get our answer. Osmotic pressure is a colligative property and thus depends upon the number of particles of a particular component and not on the nature .

Formulas used: $\pi = \dfrac{{{W_B}RT}}{{{M_B}V}}$
Where $\pi $ is the osmotic pressure, ${W_B}$ is the given mass of substance, $R$ is the universal gas constant, $T$ is the absolute temperature, ${W_B}$ is the molar mass of the substance and $V$ is the volume of the solution.
$\rho = \dfrac{{{W_B}}}{V}$
Where $\rho $ is the density.

Complete step by step answer:
The formula for calculating the osmotic pressure of a solution is given as:
$\pi = \dfrac{{{W_B}RT}}{{{M_B}V}}$
Where $\pi $ is the osmotic pressure, ${W_B}$ is the given mass of substance, $R$ is the universal gas constant, $T$ is the absolute temperature, ${W_B}$ is the molar mass of the substance and $V$ is the volume of the solution.
But as we know, the density of a substance is given as the ratio between its mass to volume. Thus, the density of the solution of the given substance will be:
$\rho = \dfrac{{{W_B}}}{V}$
Where $\rho $ is the density. Hence, we can replace the given mass term and the volume term in the osmotic pressure equation with density to get:
$\pi = \dfrac{{\rho RT}}{{{M_B}}}$
$ \Rightarrow \rho = \dfrac{{\pi {M_B}}}{{RT}}$
Here we have $\pi = 0.2atm$, ${M_B} = 240g/mol$, $R = 0.08\;litre\;atm\;{K^{ - 1}}mo{l^{ - 1}}$ and $T = 300K$. Substituting these values, we get:
$\rho = \dfrac{{0.2atm \times 240gmo{l^{ - 1}}}}{{0.08\;litre\;atm\;{K^{ - 1}}mo{l^{ - 1}} \times 300K}}$
On solving this, we get:
$\rho = 2g/litre$
But we are asked to find the density in terms of $g.d{m^{ - 3}}$
As we know, $1dm = 10cm$. Cubing both sides, we get:
$ \Rightarrow 1d{m^3} = 1000c{m^3}$
We know that $1000c{m^3} = 1litre$
Hence, we have: $1d{m^3} = 1litre = 1000c{m^3}$
Therefore, both the unit $g.d{m^{ - 3}}$ and $g/litre$ holds the same value.

Hence, density of solution $ = 2g.d{m^{ - 3}}$.

Note: Osmotic pressure is the pressure that must be applied on a given solution to prevent the movement of solvent molecules from the region of higher concentration to the region of lower concentration. Thus, when a pressure greater than the osmotic pressure is applied, the solvent molecules move from lower concentration to higher concentration region, in a phenomenon known as reverse osmosis.