Chemistry Question Bank

If at certain temperature, the vapour pressure of pure water is 25 mm Hg and that of a very dilute aqueous urea solution is 24.5 mm Hg, the molality of the solution is

Vapour pressure of dilute aqueous solution of glucose is 750 mm of Hg at 373 K. The mole fraction of solute is (vapour pressure of pure water is 760 mm of Hg) SOLUTIONS

Addition of a non-volatile solute causes lowering in vapour pressure of a solvent from 0.8 atm to 0.2 atm. What is the mole fraction of solvent?

18 g glucose (C₆H₁₂O₆) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is:

The vapour pressure of acetone at 20°C is 185 torr. When 1.2g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is:

A solution containing 12.5g of non-electrolyte substance in 185g of water shows boiling point elevation of 0.80 K. Calculate the molar mass of the substance. SOLUTIONS (Kb = 0.52 K kg mol-1)

At certain hill-station, pure water boils at 99.725ºC. If Kb for water is 0.513ºC kg mol–1, the boiling point of 0.69m solution of urea will be

An aqueous solution containing 1g of urea boils at 100.25ºC. The aqueous solution containing 3g of glucose in the same volume will boil at

If 1g of solute (molar mass = 50g mol–1) is dissolved in 50g of solvent and the elevation in boiling point is 1K. The molar boiling constant of the solvent is

The boiling point of 0.1 molal aqueous solution of urea is 100.18ºC at 1 atm. The molal elevation constant of water is

If the elevation in boiling point of a solution of 10gm of solute (mol. wt. = 100) in 100 gm of water is bT , then ebullioscopic constant of water is

Elevation in boiling point was 0.520C when 6 gm of a compound X was dissolved in 100 gm of water. Molecular weight of X is (Kb of water is 0.52 K⁻ kg/mol)

The molal b.p. constant for water is 0.5130C kg mol– 1. When 0.1 mole of sugar is dissolved in 200 g of water, the solution boils under a pressure of 1 atm at

An aqueous solution freezes at –2.550C. What is its boiling point. 0.52 / ; 1.86 / ? H O H O b f K K m K K m

Sprinkling of salt helps in clearing the snow covered roads in hills. The phenomenon involved in the process is

Given that Tf is the depression in freezing point of the solvent in a solution of a nonvolatile solute of molality 1, the quantity f m T lt m → is equal to

The freezing point of a 0.05 molal solution of a non- electrolyte in water is

Pure benzene freezes at 5.45ºC at a certain place but a 0.374m solution of tetrachloroethane in benzene freezes at 3.55ºC. The Kf for benzene is

Depression of freezing point for …(i)… is directly proportional to …(ii)… Here, (i) and (ii) refer to

What weight of glycerol should be added to 600 g of water in order to lower its freezing point by 10ºC? (Kf = 1.86ºC m–1). (Molecular mass of glycerol is 92)

The amount of urea to be dissolved in 500 cc of water (K = 1.86) to produce a depression of 0.186ºC in the freezing point is

The molar freezing point constant for water is 1.86º C/m. If 342g of cane sugar (C₁₂H₂₂O₁₁) is dissolved in 1000 g of water, the solution will freeze at

If 15 gm of a solute in 100 gm of water makes a solution that freezes at –1.0ºC, then 30 gm of the same solute in 100 gm of water will make a solution that freezes at

An aqueous solution of a non-electrolyte solute boils at 100.52ºC. The freezing point of the solution will be (Kf = 1.86K Kg mol-1, Kb = 0.52 K kg mol-1)

0.052 100.052 6 / 60 x x x x b b b b b o b T T T T T T C 102. An aqueous solution of Glucose freezes at –0.186 °C. Given that 0.512 H O b K K kg mol and 1.86 H O f K K kg mol , the elevation in boiling point of this s…