How many grams of a non-volatile solute (molar mass $40 \ g/mol$) must be added to $114 \ g$ of octane to reduce its vapour pressure by $20\%$ (in $g$)?

The vapour pressure will be lowest for

Relative lowering in vapour pressure of a solution containing a non-volatile solute is the ratio of:

Two $5 \ molal$ solutions are prepared by dissolving a non-electrolyte,non-volatile solute separately in the solvents $X$ and $Y$. The molecular weights of the solvents are $M_X$ and $M_Y$,respectively,where $M_X = \frac{3}{4} M_Y$. The relative lowering of vapour pressure of the solution in $X$ is $m$ times that of the solution in $Y$. Given that the number of moles of solute is very small in comparison to that of solvent,the value of $m$ is

At $T \ K$, $0.1 \ mol$ of a non-volatile solute was dissolved in $0.9 \ mol$ of a volatile solvent. The vapour pressure of pure solvent is $0.9 \ bar$. What is the vapour pressure (in $bar$) of the solution?

If $x \ g$ of a non-volatile,non-electrolytic solute is dissolved in $114 \ g$ of octane to reduce its vapour pressure to $80 \%$ of that of pure octane,then $x$ is ........... $g$. (Molar mass of solute $= 40 \ g/mol$)