The vapour pressure of water at room temperature is lowered by `5%` by dissolving a solute in it , then the approximately molality of solution is :

The vapour pressure of water at room temperature is lowered by 5% by dissolving a solute in it, then the approximate molality of solution is

The vapour pressure of water at room temperature is lowered by 5% by dissolving a solute in it. What is approximate molality of solution:

Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of the solution will be :

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of water in a solution of non volatile solute is 0.9, the vapour pressure of the solution will be :-

How mich urea ("molar mass"=60 g mol^(-1)) must be dissolved in 50 g of water so that the vapour pressure at room temperature is reduced by 25% ? Also calculate the molality of the solution obtained.

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

The vapour pressure of water is 12.3.kRa at 300.calculate the vapour pressure of 1 molal solution in it