Question

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 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, Δ_mixH = 0, Δ_mixV = 0]

Answer 1

Correct Answer: 682

Osmotic Pressure Calculation 

The osmotic pressure (in Torr) of the resulting solution at 300 K is \( \underline{682} \).

Mole of Urea = 0.2

Weight of Urea is: \[ 0.2 \times 60 = 12 \, \text{g} \]

Weight of solvent = 1000 g

Therefore, the volume of the solution is: \[ \frac{1012}{1.012} = 1000 \, \text{ml} \]

50 ml solution contains: \[ \frac{0.2 \times 50}{1000} = 0.01 \, \text{moles} \]

Therefore, the concentration of the solution is: \[ \frac{0.01 + 0.001}{\frac{300}{1000}} = 0.0366 \]

Finally, using the formula for osmotic pressure \( \pi = CRT \), we get: \[ \pi = 0.0366 \times 62 \times 300 = 682 \]