One mole of a non-volatile solute is dissolved in two moles of water. What is the vapor pressure of this solution relative to that of pure water?

If $P^o$ and $P$ are the vapour pressure of a solvent and its solution respectively and $N_1$ and $N_2$ are the mole fractions of the solvent and solute respectively,then the correct relation is:

Vapour pressure of $CCl_4$ at $25 \text{ }^\circ\text{C}$ is $143 \text{ mm Hg}$. If $0.5 \text{ g}$ of a non-volatile solute is dissolved in $100 \text{ cm}^3$ of $CCl_4$, find the vapour pressure of the solution. (Density of $CCl_4 = 1.58 \text{ g/cm}^3$ and molar mass of solute = $65 \text{ g/mol}$) (in $\text{ mm Hg}$)

When $18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water,the vapour pressure of the aqueous solution at $100 \ ^\circ C$ is .......... $torr$.

$60 \ g$ of Urea (Mol. wt $60$) was dissolved in $9.9 \ moles$ of water. If the vapour pressure of pure water is $P_o$,the vapour pressure of the solution is: (in $P_o$)

$1 \ mol$ 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 .......... $mm \ Hg$.