$18 \ g$ of glucose is dissolved in $90 \ g$ of water. The relative lowering of vapour pressure of the solution is equal to

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

One mole of glucose is dissolved in $2 \, moles$ of water. The vapour pressure of the solution relative to that of water is

$A$ solution is prepared by dissolving $10 \ g$ of a non-volatile solute (molar mass,'$M$ $g \ mol^{-1}$') in $360 \ g$ of water. What is the molar mass in $g \ mol^{-1}$ of solute if the relative lowering of vapour pressure of solution is $5 \times 10^{-3}$?

The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

In $108 \, g$ of water,$18 \, g$ of a non-volatile compound is dissolved. At $100^{\circ} C$,the vapour pressure of the solution is $750 \, mm \, Hg$. Assuming that the compound does not undergo association or dissociation,the molar mass of the compound in $g \, mol^{-1}$ is