The relative lowering of vapour pressure produced by dissolving $71.5 \ g$ of a substance in $1000 \ g$ of water is $0.00713$. The molecular weight of the substance will be:

The vapour pressure of pure liquids $A$ and $B$ are $450$ $mm$ $Hg$ and $700$ $mm$ $Hg$ respectively,at $350$ $K$. Find out the composition of the liquid mixture if the total vapour pressure is $600$ $mm$ $Hg$. Also,find the composition of the vapour phase.

If the vapour pressure of a solution containing a non-volatile solute is $2\%$ less than the vapour pressure of pure water,find the molality of the solution.

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

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 $mm \ Hg$)

$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)