Use the osmotic pressure relation

For dilute solutions, $\pi = C R T$ where $C$ is the total particle concentration.

Since both solutions are at the same temperature, $RT$ is common.

Find effective osmolar concentration of each solution

• For NaCl solution: $\pi_1 = 0.10\ \text{atm}$ So its effective particle concentration is proportional to $0.10$.

• For glucose solution: $\pi_2 = 0.20\ \text{atm}$ So its effective particle concentration is proportional to $0.20$.

On mixing, total osmotic pressure is based on total particles in total volume

If we mix $V_1 = 1\ \text{L}$ of NaCl solution and $V_2 = 2\ \text{L}$ of glucose solution, then $\pi_{\text{mix}} = \frac{\pi_1 V_1 + \pi_2 V_2}{V_1 + V_2}$

Substituting values: $\pi_{\text{mix}} = \frac{(0.10)(1) + (0.20)(2)}{1+2}$

$\pi_{\text{mix}} = \frac{0.10 + 0.40}{3} = \frac{0.50}{3}$

$\pi_{\text{mix}} = 0.1667\ \text{atm}$

Express in the required form

Given, $\pi_{\text{mix}} = x \times 10^{-3}\ \text{atm}$

So, $x \times 10^{-3} = 0.1667$

$x = 0.1667 \times 10^3 = 166.7$

Nearest integer: $x = 167$

Comparison with stored answer

Stored correct answer = $167$

This matches the derived answer.