When a non-volatile solute is dissolved in a solvent,the relative lowering of vapour pressure is equal to

What is the vapour pressure of a solution when $2 \ mol$ of a non-volatile solute are dissolved in $20 \ mol$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

The vapor pressure of a pure liquid solvent $(X)$ decreases from $0.80 \ atm$ to $0.60 \ atm$ upon the addition of a non-volatile solute $(Y)$. What is the mole fraction of $(Y)$ in the solution?

$6 \ g$ of a non-volatile solute $(x)$ is dissolved in $100 \ g$ of water. The relative lowering of vapour pressure of the resultant solution is $0.006$. What is the molar mass (in $g \ mol^{-1}$) of $x$?

$A$ solution was prepared by dissolving $0.1 \ mol$ of a nonvolatile solute in $0.9 \ mol$ of water. What is the relative lowering of vapour pressure of the solution?

$38.4 \ g$ of unknown substance (molar mass $384 \ g \ mol^{-1}$) and $116 \ g$ of acetone is used to prepare a solution at $313 \ K$. If vapour pressure of pure acetone (molar mass $58 \ g \ mol^{-1}$) is $0.842 \ atm$,what is the vapour pressure of the solution (in $atm$)?