Chemistry - Liquid Solution Question with Solution | TestHub

The osmotic pressure ($\pi$) of a solution is given by the formula:

$$\pi =iCRT$$

where $i$ is the van't Hoff factor, $C$ is the molar concentration of the solute, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.

For the NaCl solution:

${\pi }_1=0.10$ atm

$V_1=1$ L

So, $i_1{C}_{1}RT=0.10$ atm.

This implies $i_1{C}_1=\frac{0.10}{RT}$.

The total moles of solute particles in the NaCl solution is $n_1=i_1{C}_1{V}_1=\frac{0.10}{RT}\times 1=\frac{0.10}{RT}$.

For the glucose solution:

${\pi }_2=0.20$ atm

$V_2=2$ L

So, $i_2{C}_{2}RT=0.20$ atm.

This implies $i_2{C}_2=\frac{0.20}{RT}$.

The total moles of solute particles in the glucose solution is $n_2=i_2{C}_2{V}_2=\frac{0.20}{RT}\times 2=\frac{0.40}{RT}$.

When the two solutions are mixed, the total volume is $V_{total}=V_1+V_2=1L+2L=3L$.

The total moles of solute particles in the mixed solution is $n_{total}=n_1+n_2=\frac{0.10}{RT}+\frac{0.40}{RT}=\frac{0.50}{RT}$.

The osmotic pressure of the mixed solution, ${\pi }_{total}$, is given by:

$${\pi }_{total}=\frac{n_{total}}{V_{total}}RT$$

$${\pi }_{total}=\frac{\frac{0.50}{RT}}{3L}RT$$

$${\pi }_{total}=\frac{0.50}{3}atm$$

$${\pi }_{total}=\mathrm{0.1666...}atm$$

We are given that ${\pi }_{total}=x\times 10^{-3}$ atm.

So, $x\times 10^{-3}=\mathrm{0.1666...}$

$x=\mathrm{0.1666...}\times 10^3$

$x=\mathrm{166.6...}$

Rounding to the nearest integer, $x=167$.