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 $180 \ g \ mol^{-1}$

At $30^\circ C$,liquid $A$ and $B$ form an ideal solution. The total vapor pressure of a solution containing $1 \ mol$ of $A$ and $2 \ mol$ of $B$ is $250 \ mm \ Hg$. The total vapor pressure becomes $300 \ mm \ Hg$ when $1 \ mol$ more of $A$ is added to the first solution. What are the vapor pressures of pure $A$ and $B$ at the same temperature?

What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)

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?

$A$ solution is prepared by mixing $8.5 \ g$ of $CH_2Cl_2$ and $11.95 \ g$ of $CHCl_3$. If the vapour pressures of pure $CH_2Cl_2$ and $CHCl_3$ at $298 \ K$ are $415 \ mm \ Hg$ and $200 \ mm \ Hg$ respectively,the mole fraction of $CHCl_3$ in the vapour phase is: (Molar mass of $Cl = 35.5 \ g \ mol^{-1}$)

The vapour pressure of benzene at $90\,^oC$ is $1020 \, \text{torr}$. $5 \, g$ of a solute is dissolved in $58.5 \, g$ of benzene. The vapour pressure of the solution is $990 \, \text{torr}$. What is the molar mass of the solute?