How many grams of a non-volatile solute with a molar mass of $90$ must be dissolved in $97.5 \ g$ of water to cause a $2.5 \%$ decrease in vapour pressure?

At $298 \ K$,the vapour pressure of a solution of $7.5 \ g$ of non-volatile solute in $90 \ g$ of water is $2.8 \ kPa$. If $18 \ g$ of water is added to this solution,the vapour pressure becomes $2.81 \ kPa$ at the same temperature. The molar mass of the solute in $g \ mol^{-1}$ is:

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

Vapour pressure of a solution of benzene $(P^o_{benzene} = 120 \ torr)$ and toluene $(P^o_{toluene} = 80 \ torr)$ having $2 \ moles$ of each is ......... $torr.$

At a certain temperature,the mole fraction of a solute is $0.15$ and the vapor pressure of the pure solvent is $120 \, torr$. If the solute is non-volatile,what will be the relative lowering in vapor pressure?