50 mL of 0.2 molal urea solution (ρ = 1.012 g/mL at 300 K) is mixed with 250 mL of a solution containing 0.06 g urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in Torr) of the resulting solution at 300 K is:

[Use: Molar mass of urea = 60 g mol⁻¹; gas constant, R = 62 L Torr K⁻¹ mol⁻¹;

Assume, ΔH_mix = 0, ΔV_mix = 0]

The correct answer is 682.

Step 1: Calculate moles of urea in the first solution

The first solution has a volume $V_1=50\text{ mL}$, density $\rho =1.012\text{ g/mL}$, and molality $m=0.2\text{ mol/kg}$.

Mass of the first solution = $V_1\times \rho =50\text{ mL}\times 1.012\text{ g/mL}=50.6\text{ g}$

Let $w_1$ be the mass of solute (urea in grams). The mass of solvent in kg is given by $\frac{50.6-w_1}{1000}$.

Using the definition of molality:

$m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}=\frac{w_1/60}{(50.6-w_1)/1000}=0.2$

$\frac{1000w_1}{60(50.6-w_1)}$

$1000w_1=12(50.6-w_1)=607.2-12w_1$

$1012w_1=607.2\Rightarrow w_1=0.6\text{ g}$

Moles of urea in the first solution, $n_1=\frac{0.6}{60}=0.01\text{ mol}$.

Step 2: Calculate moles of urea in the second solution

The second solution contains $0.06\text{ g}$ of urea in $V_2=250\text{ mL}$.

Moles of urea in the second solution, $n_2=\frac{0.06}{60}=0.001\text{ mol}$.

Step 3: Calculate the total moles of urea and total volume after mixing

Total moles of solute, $n_{\text{total}}=n_1+n_2=0.01+0.001=0.011\text{ mol}$.

Since $\Delta V_{\text{mix}}=0$, the total volume of the solution is:

$V_{\text{total}}=V_1+V_2=50\text{ mL}+250\text{ mL}=300\text{ mL}=0.3\text{ L}$.

Step 4: Calculate the osmotic pressure (π) of the resulting solution

Using the formula for osmotic pressure $\pi =CRT=\frac{n_{\text{total}}}{V_{\text{total}}}RT$:

Given parameters:

• $R=62\text{ L Torr K}{}^{-1}\text{mol}{}^{-1}$

• $T=300\text{ K}$

$\pi =\frac{0.011\text{ mol}}{0.3\text{ L}}$

$\pi =0.011\times 62\times 1000=11\times 62=682\text{ Torr}$

Thus, the osmotic pressure of the resulting solution is 682 Torr.