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

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 , then the approximately molality of solution is :

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.