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 pressures of two volatile liquids $A$ and $B$ at $25^{\circ} C$ are $50 \ Torr$ and $100 \ Torr,$ respectively. If the liquid mixture contains $0.3$ mole fraction of $A$,then the mole fraction of liquid $B$ in the vapour phase is $\frac{x}{17}$. The value of $x$ is $...$

$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

The vapour pressure of an aqueous solution of methanol is:

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 $mm \ Hg$)

At $300 \ K$,the vapour pressure of a solution containing $1 \ mole$ of $n-hexane$ and $3 \ moles$ of $n-heptane$ is $550 \ mm \ Hg$. At the same temperature,if one more mole of $n-heptane$ is added to this solution,the vapour pressure of the solution increases by $10 \ mm \ Hg$. What is the vapour pressure in $mm \ Hg$ of $n-heptane$ in its pure state?