What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water,what is the vapour pressure (in $mm$) of the resultant solution?

At $300 \ K$,the vapour pressure of an ideal solution containing $3 \ mol$ of $A$ and $2 \ mol$ of $B$ is $600 \ torr$. At the same temperature,if $1.5 \ mol$ of $A$ and $0.5 \ mol$ of $C$ (non-volatile) are added to this solution,the vapour pressure of the solution increases by $30 \ torr$. What is the value of $p_B^o$?

Which one of the following represents the graph between $\log p$ (on $Y$-axis) and $\frac{1}{T}$ (on $X$-axis)?
($p=$ vapour pressure of a liquid,$T=$ absolute temperature)

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 $mm \ Hg$)

$1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ is mixed with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$. The vapor pressure of the resulting ideal solution is .......... $mm \ Hg$.