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 .............. $g$.

The vapour pressure of water at $20\,^{\circ}C$ is $17.54\,mm$. When $20\,g$ of a non-ionic substance is dissolved in $100\,g$ of water,the vapour pressure is lowered by $0.30\,mm$. What is the molecular weight of the substance?

Calculate the relative lowering of vapour pressure of a solution containing $0.56 \ g$ of a non-volatile solute in $100 \ g$ of water [molar mass of solute $= 60 \ g \ mol^{-1}$].

The vapour pressure of pure benzene at a certain temperature is $640 \ mm \ Hg$. $A$ non-volatile non-electrolyte solid weighing $2.175 \ g$ is added to $39.0 \ g$ of benzene. The vapour pressure of the solution is $600 \ mm \ Hg$. What is the molar mass of the solid substance?

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 water. The relative lowering of vapour pressure is