The vapour pressure of a solvent decreases by $10 \ mm \ Hg$ if the mole fraction of a non-volatile solute is $0.2$. Calculate the vapour pressure of the pure solvent. (in $mm \ Hg$)

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water,what is the vapour pressure (in $mm$) of the resultant solution?

The mass of glucose that should be dissolved in $100 \ g$ of water in order to produce the same relative lowering of vapour pressure as is produced by dissolving $1 \ g$ of urea (molar mass $= 60 \ g/mol$) in $50 \ g$ of water is ........... $g$. (Assume dilute solution in both cases):-

$3 \ g$ of urea is dissolved in $45 \ g$ of $H_2O$. The relative lowering in vapour pressure is

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

What is the vapour pressure of a solution containing $0.1 \ mol$ of non-volatile solute dissolved in $16.2 \ g$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$