What is the relation between the vapour pressure of the solution,the vapour pressure of the pure solvent,and its mole fraction in the solution?

$3 \ g$ of urea is dissolved in $45 \ g$ of water. The relative lowering of vapour pressure is

At $298 \ K$,the vapour pressure of pure water is $25 \ torr$. The vapour pressure of water,when $12 \ g$ of urea (molar mass,$60 \ g \ mol^{-1}$) and $36 \ g$ of glucose (molar mass,$180 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of water at the same temperature (in $torr$) is

The vapour pressure of a pure solvent is $0.70 \ atm.$ When a non-volatile solute $A$ is dissolved in it,the vapour pressure of the solution becomes $0.35 \ atm.$ The mole fraction of solute $A$ will be$:-$

$12 \ g$ of urea is dissolved in $1 \ L$ of water and $68.4 \ g$ of sucrose is dissolved in $1 \ L$ of water. The lowering of vapour pressure of the first case is

The vapour pressure of pure benzene at a certain temperature is $0.850 \ bar$. $A$ non-volatile,non-electrolyte solid weighing $0.5 \ g$ when added to $39.0 \ g$ of benzene (molar mass $78 \ g \ mol^{-1}$),the vapour pressure of the solution becomes $0.845 \ bar$. What is the molar mass of the solid substance?