Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are M_x and M_y, respectively where M_x = $${3 \over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :

A solution is prepared by mixing 8.5 g of CH₂Cl₂ and 11.95 g of CHCl₃ . If vapour pressure of CH₂Cl₂ and CHCl₃ at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl₃ in vapour form is : (Molar mass of Cl = 35.5 g mol⁻¹)

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82^oC. If Na₂SO₄ is 81.5% ionised, the value of x

(K_f for water=1.86^oC kg mol⁻¹) is approximately :

(molar mass of S = 32 g mol⁻¹ and that of Na = 23 g mol⁻¹)

The freezing point of benzene decreases by 0.45⁰C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (K_f for benzene = 5.12 K kg mol^–1)