Sea water contains 29 . 25 % NaCl and 19 % MgCl 2 by weight of solution. The normal boiling point of the sea…

Molality of solution$=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solvent}}\times 1000$

Number of moles $\mathrm{NaCl} =\frac{29.25}{58.5}=0.5$

Number of moles of ${\mathrm{MgCl}}_2$$=\frac{19}{95}=0.2$

$\mathrm{NaCl}=\frac{0.5}{51.75}\times 1000$

${\mathrm{MgCl}}_2=\frac{0.2}{51.75}\times 1000$

$\left({\mathrm{\Delta T}}_{\mathrm{b}}\right)=\left\{\left({\mathrm{i}}_1 {\mathrm{m}}_1\right)+\left({\mathrm{i}}_2 {\mathrm{m}}_2\right)\right\}{\mathrm{k}}_{\mathrm{b}}$

$∆{\mathrm{T}}_{\mathrm{b}}$= Elevation in boiling point

$\mathrm{i}=$Van't Hoff factor

m = molality

${\mathrm{k}}_{\mathrm{b}}$ = molal elevation constant

$∆{\mathrm{T}}_{\mathrm{b}=}\left(\frac{2\times 0.5\times 1000}{51.75}\times \frac{3\times 0.2\times 1000}{51.75}\right)\times 0.52$

$= 16.077$

Boiling point of sea water

$=116.077°\mathrm{C}$

$\approx 116°\mathrm{C}$ (Nearest integer)