The vapour pressure of a given liquid decreases if?

Calculate the vapour pressure of a solution containing a mixture of $2 \ moles$ of volatile liquid $A$ and $3 \ moles$ of volatile liquid $B$ at room temperature. $(P_{A}^{\circ} = 420 \ mm \ Hg, P_{B}^{\circ} = 610 \ mm \ Hg)$ (in $mm \ Hg$)

If two substances $A$ and $B$ have $P_A^0 : P_B^0 = 1 : 2$ and have mole fraction in solution $1 : 2$,then what is the mole fraction of $A$ in the vapour phase?

Two liquids $A$ and $B$ form a solution at $306 \ K$ that follows the equation $P \ (\text{in } atm) = 0.172X_A + 0.215$,where $P$ is the total vapor pressure of the solution and $X_A$ is the mole fraction of $A$. The vapor pressure of pure $B$ is ............ . (in $atm$)

The vapour pressure of pure benzene at a certain temperature is $200 \ mm \ Hg$. At the same temperature,the vapour pressure of a solution containing $2 \ g$ of a non-volatile,non-electrolyte solid in $78 \ g$ of benzene is $195 \ mm \ Hg$. What is the molecular weight of the solid?

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