At a temperature of $300 \ K$,if a liquid is transferred from a small test tube to a large beaker,what will be the change in its vapor pressure?

Consider a binary solution of two volatile liquid components $1$ and $2$. $x_1$ and $y_1$ are the mole fractions of component $1$ in liquid and vapour phase,respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as :

The vapour pressure of a solvent decreases by $10 \ mm \ Hg$ if the mole fraction of a non-volatile solute is $0.2$. Calculate the vapour pressure of the pure solvent. (in $mm \ Hg$)

At $T$ $(K)$,the vapour pressure of pure benzene is $0.85$ bar. $A$ non-volatile,non-electrolyte substance weighing $0.5$ $g$ when added to $39$ $g$ of benzene,the vapour pressure of the solution is $0.845$ bar. The molar mass (in $g$ $mol^{-1}$) of the substance is

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

$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is: