According to Raoult's law,the relative lowering of vapour pressure of a solution of a non-volatile solute is equal to:

Dry air was drawn through bulbs containing pure water,then through bulbs containing a solution of $600 \ g$ of a non-electrolyte in $360 \ g$ of water at the same temperature,and finally through a tube in which dried $CaCl_2$ was placed. The solution bulb gained $1.5 \ g$ and the dried $CaCl_2$ gained $2 \ g$. The molecular mass of the solute is:

$A$ 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.

The correct option for the value of vapour pressure of a solution at $45^{\circ} C$ with benzene to octane in molar ratio $3 : 2$ is ...... $mm$ of $Hg$. [At $45^{\circ} C$,vapour pressure of benzene is $280 \ mm \ Hg$ and that of octane is $420 \ mm \ Hg$. Assume ideal solution.]

The vapour pressure of benzene at a certain temperature is $640 \, mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \, g$ is added to $39.08 \, g$ of benzene. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?

Vapour pressure of a solution and of a pure solvent are $P_1$ and $P_1^0$ respectively. If $\frac{P_1}{P_1^0}$ is $0.15$,find the mole fraction of the solute.