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:

Calculate the weight of non-volatile solute having molecular weight $40$,which is dissolved in $57 \ g$ octane to reduce its vapour pressure by $20\%$.

Calculate the vapour pressure of the solution if the relative lowering of vapour pressure and the vapour pressure of the pure solvent are $0.018$ and $18 \ mm \ Hg$ respectively at $300 \ K$. (in $mm \ Hg$)

$A$ solution is made by mixing $1 \ mole$ of volatile liquid $A$ with $3 \ moles$ of volatile liquid $B$. The vapour pressure of pure $A$ is $200 \ mm \ Hg$ and that of the solution is $500 \ mm \ Hg$. The vapour pressure of pure $B$ and the least volatile component of the solution,respectively,are $:$

The relative lowering of vapour pressure produced by dissolving $18 \text{ g}$ of urea (molar mass = $60 \text{ g mol}^{-1}$) in $100 \text{ g}$ of water is

The mass of non-volatile,non-electrolyte solute (molar mass $= 50 \ g \ mol^{-1}$) needed to be dissolved in $114 \ g$ octane to reduce its vapour pressure to $75\%$ is .............. $g$.