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

Urea Solution and Osmotic Pressure Calculation

Mole of Urea = 0.2

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

Weight of solvent = 1000 g

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

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

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

Using the formula for osmotic pressure: \[ \pi = C R T = 0.0366 \times 62 \times 300 = 682 \]

Final Answer:

The osmotic pressure \( \pi \) is 682.

Answer 2

1. Given:
   • Mole of urea = 0.2
   • Weight of urea = 0.2×60=12 g
   • Weight of solvent = 1000 g
   • Weight of solution = 1012 g
2. Calculations:
   • Volume of solution = \(\frac{1012}{1.012} = 1000 \, \text{ml}\)
   • Mole of urea in 50 ml solution = \(0.2 \times \frac{50}{1000} = 0.01\)
   • Mole of urea in other solution = \(\frac{0.06}{60} = 0.001\)
   • Concentration of solution = \(0.01 + \frac{0.001}{300} = 0.0366\)
   • Osmotic pressure \(\pi = C \cdot R \cdot T = 0.0366 \times 62 \times 300 = 682\)

So, the osmotic pressure of the solution is 682 Torr.