50 mL of 0.2 molal urea solution (density = 1.012 g mL$$^{-1}$$ at 300 K) is mixed with 250 mL of a solution containing 0.06 g of 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, $$\Delta_{mix}H = 0$$, $$\Delta_{mix}V = 0$$]

Osmotic pressure is given by $$\pi=\dfrac{n_{\text{total}}RT}{V_{\text{solution}}}$$, where
$$n_{\text{total}}$$ = total number of moles of solute after mixing,
$$R = 62\;{\text{L Torr K}}^{-1}{\text{mol}}^{-1}$$ (given),
$$T = 300\;\text{K}$$ (given),
$$V_{\text{solution}}$$ = final volume of the mixed solution in litres.

Step 1: Moles of urea in the 50 mL, 0.2 molal solution

Molality definition: $$m = \dfrac{n}{\text{mass of solvent (kg)}}$$

Let the mass of solvent in this solution be $$w\; \text{g}$$.
Then $$n = m \times \dfrac{w}{1000} = 0.2 \times \dfrac{w}{1000}=0.0002w$$ mol.

Mass of solute $$= n \times M = 0.0002w \times 60 = 0.012w \;\text{g}$$.

Total mass of this solution $$= w + 0.012w = 1.012w \;\text{g}$$.

Using the density (1.012 g mL$$^{-1}$$):
$$\text{mass} = \text{density} \times \text{volume} = 1.012 \times 50 = 50.6 \;\text{g}$$.

So $$1.012w = 50.6 \;\Rightarrow\; w = 50.0\;\text{g}$$.

Therefore, moles of urea in this portion
$$n_1 = 0.0002 \times 50 = 0.01\;\text{mol}$$.

Step 2: Moles of urea in the 250 mL second solution

Given mass of urea = 0.06 g.
$$n_2 = \dfrac{0.06}{60} = 0.001\;\text{mol}$$.

Step 3: Total moles after mixing

$$n_{\text{total}} = n_1 + n_2 = 0.01 + 0.001 = 0.011\;\text{mol}$$.

Step 4: Total volume of the mixed solution

Assuming $$\Delta_{mix}V = 0$$ (volume is additive):
$$V_{\text{solution}} = 50\;\text{mL} + 250\;\text{mL} = 300\;\text{mL} = 0.300\;\text{L}$$.

Step 5: Osmotic pressure

$$\pi = \dfrac{n_{\text{total}} RT}{V_{\text{solution}}} = \dfrac{0.011 \times 62 \times 300}{0.300}$$

First compute the concentration:
$$\dfrac{0.011}{0.300}=0.0366667\;\text{mol L}^{-1}$$.

Then
$$\pi = 0.0366667 \times 62 \times 300 = 0.0366667 \times 18\,600 \approx 682\;\text{Torr}$$.

Hence, the osmotic pressure of the resulting solution at 300 K is 682 Torr.