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

Mole of Urea = 0.2
Weight of Urea = \(0.2\times60=12g\)
Weight of solvent = 1000 g
\(\therefore\) The volume of solution = \(\frac{1012}{1.012}=1000\,\,ml\)
\(\therefore\) 50 ml solution contains = \(\frac{0.2\times50}{1000}=0.01\)
\(\therefore\) The concentration of solution = \(\frac{0.01+0.001}{\frac{300}{1000}}\) = 0.0366
\(\therefore \, \pi=CRT=0.0366\times62\times300=682\)

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.