After adding a non-volatile solute, the freezing point of water decreases to $-0.186^{\circ} C$. Calculate $\Delta T_b$ if $K_f = 1.86 \text{ K kg mol}^{-1}$ and $K_b = 0.521 \text{ K kg mol}^{-1}$. (in $\text{ K}$)

At $80^o C$,the vapor pressure of pure liquid $A$ is $520 \ mm \ Hg$ and that of pure liquid $B$ is $1000 \ mm \ Hg$. If a mixture of $A$ and $B$ boils at $80^o C$ and $1 \ atm$ pressure,the mole percentage of $A$ in the mixture is ........... $(1 \ atm = 760 \ mm \ Hg)$.

For a dilute solution containing $2.5 \ g$ of a non-volatile non-electrolyte solute in $100 \ g$ of water,the elevation in boiling point at $1 \ atm$ pressure is $2^{\circ} C$. Assuming the concentration of solute is much lower than the concentration of solvent,the vapour pressure ($mm$ of $Hg$) of the solution is (take $K_{b}=0.76 \ K \ kg \ mol^{-1}$)

If the freezing point of an aqueous solution of salt $MX_4$ $(\alpha = 90\%)$ is $-3\ ^oC$,then its boiling point will be ........ $^oC$. (Assume it is $100\%$ ionized at the boiling temperature and $k_f = 1.86\ K\ kg\ mol^{-1}$,$k_b = 0.52\ K\ kg\ mol^{-1}$)

Select the correct statement for an ideal solution of a non-volatile solute in a liquid solvent.

An aqueous solution freezes at $-0.186 \ ^\circ C$ ($K_f = 1.86 \ ^\circ C \ kg \ mol^{-1}$; $K_b = 0.512 \ ^\circ C \ kg \ mol^{-1}$). What is the elevation in boiling point of the same solution?