Sea water contains 29 . 25 % NaCl and 19 % MgCl _ 2 by weight of...

Molality of solution$= \dfrac{molesofsolute}{massofsolvent} \times 1000$

Number of moles $NaCl = \dfrac{29 . 25}{58 . 5} = 0 . 5$

Number of moles of ${MgCl}_{2}= \dfrac{19}{95} = 0 . 2$

$NaCl = \dfrac{0 . 5}{51 . 75} \times 1000$

${MgCl}_{2} = \dfrac{0 . 2}{51 . 75} \times 1000$

\left({\Delta T}_{b}\right) = \left\{\left({i}_{1}{m}_{1}\right) + \left({i}_{2}{m}_{2}\right)\right\}{k}_{b}$</p><p>$∆ {T}_{b}$= Elevation in boiling point</p><p>$i =$Van't Hoff factor</p><p>m = molality</p><p>${k}_{b}$= molal elevation constant</p><p>$∆ {T}_{b =}\left(\dfrac{2 \times 0 . 5 \times 1000}{51 . 75} \times \dfrac{3 \times 0 . 2 \times 1000}{51 . 75}\right) \times 0 . 52$</p><p>$= 16 . 077$</p><p>Boiling point of sea water</p><p>$= 116 . 077 ° C$</p><p>$\approx 116 ° C (Nearest integer)