Vapour pressure of benzene is 200 mm Hg . When 2 g of non-volatile solute is dissolved in 78 g benzene, benzene solution has vapour pressure 195 mm Hg . Calculate the molar mass of the solute in gmol - 1 ( molar mass of benzene = 78 gmol - 1 ).

Given, P solution = 195 mm Hg P solvent = 200 mm Hg Using Raoult's law, ⇒ P solution = P solvent × χ solvent ⇒ P solution = P solvent × n solvent n solvent + n solute ⇒ 195 = 200 × 1 1 + 2 M solute ⇒ M solute = 80 gmol - 1 Therefore, the molar mass of the solute is 80 g / mol - 1 .

What is the value of Van’t Hoff factor for K 2 SO 4 ?

What is the value of Van’t Hoff factor for K 2 SO 4 is i = 3 . It is a strong electrolyte and dissociates its ions very easily in water. Therefore, the chemical reaction observed is- K 2 SO 4 → 2 K + + SO 4 2 - i = 2 + 1 = 3

Give reasons for the following: Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.

KCl being an electrolyte dissociates almost completely to give K + and Cl − ions whereas glucose being non-electrolyte does not dissociate. Hence, the number of particles in 1 M KCl solution is nearly double that in the 1 M sugar solution. Elevation of boiling point, being a colligative property (these properties depend on the number of solute particles) is, therefore, nearly twice for KCl solution than for glucose solution of same molarity.

Vapour pressure of water at 293 K is 17 . 535 mm Hg . Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Given:- P 0 = 17 . 535 mm Molar mass of glucose ( C 6 H 12 O 6 ) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1 Molar mass of water = 18 g mol - 1 Raoult ' s law p 0 - p s p 0 = n 2 n 1 + n 2 = n 2 n 1 = w 2 / M 2 w 1 / M 1 1 - p s p 0 = 25 / 180 450 / 18 = 25 × 18 180 × 450 1 - 450 81000 = p s p 0 p s = 0 . 9945 × 17 . 535 = 17 . 44 mm Hg