1. Find the composition of the first solution

Given:

• Volume of solution $=50\,\text{mL}$
• Density $=1.012\,\text{g mL}^{-1}$
• Molality $=0.2\,\text{mol kg}^{-1}$

So, mass of first solution: $m_{\text{solution,1}}=50\times 1.012=50.6\,\text{g}$

Let mass of solvent in this solution be $w$ g. Then mass of urea is: $\text{moles of urea}=0.2\times \frac{w}{1000}=0.0002w$ $\text{mass of urea}=60\times 0.0002w=0.012w\,\text{g}$

Now, $w+0.012w=50.6$ $1.012w=50.6$ $w=50\,\text{g}$

Hence, solvent mass $=50$ g and urea mass: $0.012\times 50=0.6\,\text{g}$

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

2. Find moles of urea in the second solution

Given $0.06$ g urea in $250$ mL solution. So, $n_2=\frac{0.06}{60}=0.001\,\text{mol}$

3. After mixing

Since $\Delta V_{\text{mix}}=0$, total volume is additive: $V_{\text{total}}=50+250=300\,\text{mL}=0.300\,\text{L}$

Total moles of urea: $n=n_1+n_2=0.01+0.001=0.011\,\text{mol}$

Molarity of resulting solution: $C=\frac{0.011}{0.300}=0.036666\ldots\,\text{mol L}^{-1}$

4. Calculate osmotic pressure

For urea, $i=1$.

$\pi = CRT$ $\pi=(0.036666\ldots)(62)(300)$

Now, $62\times 300=18600$ $\pi=0.036666\ldots\times 18600=682\,\text{Torr}$

5. Final answer

$\boxed{682}$

The derived answer matches the stored correct answer.