Calculate the normal boiling point of a sample of sea water found to contain 3.5% of NaCl and 0.13% of ${\mathrm{MgCl}}_2$ by mass. The normal boiling point of water is ${100}^{o}C$ and ${\mathrm{K}}_{\mathrm{b}}$ (water) = 0.51 K kg ${\mathrm{mol}}^{-1}$. Assume that both the salts are completely ionised

Mass of NaCl = 3.5 g

Number of moles of $\mathrm{NaCl}=\frac{3.5}{58.5}$

Number of ions furnished by one molecule of NaCl is 2. 

So, actual number of moles of particles furnished by sodium chloride $=2\times \frac{3.5}{58.5}$

Similarly, actual number of moles of particles furnished by magnesium chloride $=3\times \frac{0.13}{95}$

Total number of moles of particles $=(2\times \frac{3.5}{58.5}+3\times \frac{0.13}{95})=0.1238$

Mass of water $=(100-3.5-0.13)=96.37g=\frac{96.37}{1000}\text{ kg}$

Molality $=\frac{0.1238}{96.37}\times 1000=1.2846$

${\mathrm{\Delta T}}_{\mathrm{b}}=$Molality $\times {\mathrm{K}}_{\mathrm{b}}=1.2846\times 0.51=0.655\mathrm{K}$

Hence, boiling point of sea water = 373.655 K or ${100.655}^{o}C$