The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is 0.20 atm. The osmotic pressure of a solution formed by mixing 1 L of the sodium chloride solution with 2 L of the glucose solution is $$x \times 10^{-3}$$ atm. x is __________ (nearest integer)

We recall the van’t Hoff expression for osmotic pressure: $$\pi = iCRT,$$ where $$\pi$$ is the osmotic pressure, $$i$$ is the van’t Hoff factor, $$C$$ is the molar concentration of the solute particles, $$R$$ is the gas constant and $$T$$ is the absolute temperature.

Because all the solutions are at the same temperature and will be mixed at that temperature, the product $$RT$$ will remain the same throughout our calculations. This lets us treat $$\pi$$ as being directly proportional to the “osmolarity” (the effective concentration of dissolved particles). In other words, for any solution,

$$\pi = (iC)RT \quad\Longrightarrow\quad iC = \dfrac{\pi}{RT}.$$

Hence the quantity $$\dfrac{\pi}{RT}$$ can be viewed as the effective number of moles of solute particles per litre. If a solution of volume $$V$$ litres has osmotic pressure $$\pi,$$ then the total effective moles of solute particles present in that volume will be

$$n_{\text{eff}} = \left(\dfrac{\pi}{RT}\right)V.$$

We now apply this idea separately to the two solutions being mixed.

For the NaCl solution
Given: $$\pi_1 = 0.10\;\text{atm}, \; V_1 = 1\;\text{L}$$

Effective moles of solute particles:

$$n_{\text{eff,1}} = \left(\dfrac{\pi_1}{RT}\right)V_1 = \left(\dfrac{0.10}{RT}\right)\times 1.$$

For the glucose solution
Given: $$\pi_2 = 0.20\;\text{atm}, \; V_2 = 2\;\text{L}$$

Effective moles of solute particles:

$$n_{\text{eff,2}} = \left(\dfrac{\pi_2}{RT}\right)V_2 = \left(\dfrac{0.20}{RT}\right)\times 2.$$

Adding the two solutions
Total effective moles of solute particles after mixing:

$$n_{\text{eff,total}} = n_{\text{eff,1}} + n_{\text{eff,2}} = \left(\dfrac{0.10}{RT}\right)\times 1 + \left(\dfrac{0.20}{RT}\right)\times 2 = \dfrac{0.10 + 0.40}{RT} = \dfrac{0.50}{RT}.$$

Total volume after mixing:

$$V_{\text{total}} = V_1 + V_2 = 1\;\text{L} + 2\;\text{L} = 3\;\text{L}.$$

Osmotic pressure of the mixture
We again use $$\pi = (iC)RT = \dfrac{n_{\text{eff,total}}}{V_{\text{total}}}RT.$$ Substituting the values obtained above gives

$$\pi_{\text{mix}} = \dfrac{n_{\text{eff,total}}}{V_{\text{total}}}RT = \dfrac{\left(\dfrac{0.50}{RT}\right)}{3}\,RT = \dfrac{0.50}{3} = 0.166666\ldots\;\text{atm}.$$

Expressing this in the requested form $$x \times 10^{-3}\;\text{atm}$$:

$$0.166666\;\text{atm} \approx 1.66666 \times 10^{-1}\;\text{atm} = 166.666 \times 10^{-3}\;\text{atm}.$$

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

So, the answer is $$167.$$