A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass 60 g mol-1 and 0.9 g of a non-volatile non-electrolyte solute 'B' of molar mass 180 g mol-1 in 100 mL H2O at 27 . Osmotic pressure of the solution will be [Given: R = 0.082 L atm K-1 mol-1]
- 2.46 atm
- 0.82 atm
- 1.47 atm
- 1.23 atm
Answer
(1) 2.46 atm
Solution
Osmotic pressure depends on the total number of solute particles: = (nA + nB)VRT
nA = 0.360 = 0.005 mol, nB = 0.9180 = 0.005 mol
nA + nB = 0.01 mol; V = 100 mL = 0.1 L; T = 300 K
= 0.010.1 0.082 300
= 0.1 24.6 = 2.46 atm
Answer: (1)