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

$A$ solution is obtained by dissolving $12 \ g$ of urea (mol.wt. $60$) in a litre of water. Another solution is obtained by dissolving $68.4 \ g$ of cane sugar (mol.wt. $342$) in a litre of water at the same temperature. The lowering of vapour pressure in the first solution is

Calculate the relative lowering of vapour pressure of a solution containing $46 \ g$ of non-volatile solute in $162 \ g$ of water at $20^{\circ} C$. [Molar mass of non-volatile solute $= 46 \ g \ mol^{-1}$]

$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

The vapour pressure of $30 \%$ $(w/v)$ aqueous solution of glucose is $...... \ mm \ Hg$ at $25^{\circ} \ C$. [Given : The density of $30 \%$ $(w/v)$ aqueous solution of glucose is $1.2 \ g \ cm^{-3}$ and vapour pressure of pure water is $24 \ mm \ Hg$.] (Molar mass of glucose is $180 \ g \ mol^{-1}$.)