QJEEAdv23Chc084: JEE Advanced 2023 Chemistry - Class 12 JEEAdv23Chc084 Solution: of 0.2 molal urea solution (density at ) is mixed with of a…

Find the weight of urea in the 0.2 molal solution :

Given a 0.2 molal solution, there are 0.2 moles of urea in 1000 g of solvent. The weight of urea is thus $0.2 \, \text{mol} \times 60 \, \text{g/mol} = 12 \, \text{g}$ .

Find the weight of the solution :

The total weight of the solution is the weight of the solvent plus the weight of the solute, or $1000 \, \text{g} + 12 \, \text{g} = 1012 \, \text{g}$ .

Find the volume of the solution :

Given the density of the solution, we can find its volume by dividing the total weight of the solution by the density: $\frac{1012 \, \text{g}}{1.012 \, \text{g/mL}} = 1000 \, \text{mL}$ .

Find the amount of urea in 50 mL of the 0.2 molal solution :

If 1000 mL of the solution contains 0.2 moles of urea, then 50 mL of the solution contains $\frac{0.2 \, \text{mol} \times 50 \, \text{mL}}{1000 \, \text{mL}} = 0.01 \, \text{mol}$ of urea.

Find the amount of urea in the 250 mL solution :

The 250 mL solution contains $0.06 \, \text{g}$ of urea, or $\frac{0.06 \, \text{g}}{60 \, \text{g/mol}} = 0.001 \, \text{mol}$ .

Find the total concentration of the solution :

After mixing, the total volume of the solution is $50 \, \text{mL} + 250 \, \text{mL} = 300 \, \text{mL}$ , and the total amount of urea is $0.01 \, \text{mol} + 0.001 \, \text{mol} = 0.011 \, \text{mol}$ .

So, the concentration of the solution is $\frac{0.011 \, \text{mol}}{300 \, \text{mL}} \times 1000 \, \text{mL/L} = 0.0366 \, \text{M}$ .

Find the osmotic pressure of the solution :

Finally, the osmotic pressure $\pi$ of the solution can be found using the formula $\pi = CRT$ , where $C$ is the concentration, $R$ is the gas constant, and $T$ is the temperature. Substituting the given and calculated values,

we have $\pi = 0.0366 \, \text{M} \times 62 \, \text{L Torr K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} = 682 \, \text{Torr}$ .