2molal aqueous solution of an electrolyte  $X_2{Y}_3$ is 75% ionized. The boiling point of the solution at 1atm is  $[K_b\left(H_{2}O\right)=0.52Kkg/mol]$

$X_2{Y}_3\to 2X^{+3}+3Y^{2-}$
$n=5$
$\alpha =\frac{i-1}{n-1}\Rightarrow 0.75=\frac{i-1}{5-1}\Rightarrow i=4\left(0.75\right)+1$
$=3.0+1$
$i=4$
$\Delta T_b=i\times K_b\times m$
$=4\times 0.52\times 2$
$=4.16K$
$\therefore \Delta T_b={\left(T_b\right)}_{so\ln }-\left(T_b\right)°$
$\therefore {\left(T_b\right)}_{so\ln }=\Delta T_b+T_b°$
$=373+4.16$
$=377.16K$