For two components $A$ and $B$,$P_A^0 : P_B^0 = 1:2$ and $X_A : X_B = 1:2$. The mole fraction of $A$ in the vapor phase is:

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 .......... $torr$.

$A$ solution containing $30 \,g$ of non-volatile solute in $90 \,g$ of water has a vapour pressure of $2.8 \,kPa$ at $298 \,K$. Further,$18 \,g$ of water is added to the solution and the new vapour pressure becomes $2.9 \,kPa$ at $298 \,K$. Calculate:
$i$. Molar mass of the solute
$ii$. Vapour pressure of water at $298 \,K$

What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water $\left(P_1^0 = 32 \ mm \ Hg\right) (in $mm \ Hg$)?$

Assertion $:-$ Vapour pressure of a liquid increases on increasing surface area.
Reason $:-$ Rate of vaporisation of a liquid is directly proportional to surface area.

$A$ graph of vapour pressure and temperature for three different liquids $X$,$Y$ and $Z$ is shown below.
The following inferences are made:
$(A)$ $X$ has higher intermolecular interactions compared to $Y$.
$(B)$ $X$ has lower intermolecular interactions compared to $Y$.
$(C)$ $Z$ has lower intermolecular interactions compared to $Y$.
The correct inference$(s)$ is/are: