Answer
$104.6C^{\circ}$
Work Step by Step
We can find the required boiling point as follows:
$n_{solute}=23.8mol-n_{H_2O}$
$n_{solute}=23.8mol-20.5mol=3.3mol$
Now the molarity of the solution is
$m=(\frac{2.3mol\space solute}{20.5mol\space solv})(\frac{1mol\space solv}{18.0152g\space solv})(\frac{1000g\space solv}{1Kg\space solv})=8.9355m$
$\Delta T_b=K_b\times m$
$\Delta T_b=(0.512\frac{C^{\circ}}{m})(8.9355m)=4.6C^{\circ}$
$boiling \space point=100.0C^{\circ}+\Delta T_b$
We plug in the known values to obtain:
$boiling \space point=100.0C^{\circ}+4.6C^{\circ}=104.6C^{\circ}$
