A solution is prepared by dissolving 0.6 g of urea (molar mass $= 60 g m o l^{- 1}$ ) and $1.8 g$ of glucose (molar mass $= 180 g m o l^{- 1}$ ) in 100 mL of water at $27^{o} C .$ The osmotic pressure of the solution is:
( $R = 0.08206 L$ $atm$ $K^{- 1} m o l^{- 1}$ )

Question

TestHub · Accepted Answer

i factor of glucose and urea are 1 : 1 respectively. Moles of glucose m o l ec u l a r w e i g h t ma ss ​ = 180 1.8 ​ = 0.01 Moles of urea = 0.6 60 = 10 - 2 Total mole of solute = 10 - 2 + 10 - 2 = 2 × 10 - 2 Concentration of solution = 2 × 10 - 2 100 × 10 - 3 = 0.2 Osmotic pressure ( π ) = C R T = 0.2 × 0.0821 × 300 = 4.926 a t m

Chemistry - Liquid Solution Question with Solution | TestHub

Options:

Doubts & Discussion