Dry air was successively passed through a solution of 5 g solute in 80 g water & then through pure water. The loss in weight of solution was 2.5 g & that of pure water was 0.04 g. What is the mol. wt. of solute ?

Dry air was passed successively through a solution of 5 gm of solute dissolved in 80.0 gm of water and through pure water. The loss in weight of the solution was 2.5 gm and that of the pure solvent was 0.04 gm. What is the molecular weight of the solute?

SO_(2) gas is passed through a clear solution of lime water.

Addition of a solute to pure water causes

Water potential of pure water and its solution are

Water potentials of pure water and its solution are

The weight of a body in air is 100 g and its weight in water is 40 g. Find the volume of the body.

A solution contains 5.6 gm of glucose in 1000 gm of water. Calculate the boiling point of the solution. ( k_b for water = 0.52 K kg mol^-1 )

At 100^@C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K_b = 0. 52, the boiling point of this solution will be.

At 100^(@)C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If k_(b) = 0.5 , the boiling point of this solution will be -

A solution is prepared by dissolving 10 g of nonvolatile solute in 180 g of H_2O . If the relative lowering of vapour pressure is 0.005, find the mol. wt. of the solute.