1. Tardigrade
  2. Question
  3. Chemistry
  4. Sea water is found to contain 5.85 % NaCl and 9.50 % MgCl 2 be weight of solution. Calculate its normal boiling point assuming 80 % ionisation for NaCl and 50 % ionisation of MgCl 2[ K b( H 2 O ). =0.51 kg mole -1 K ].

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Q. Sea water is found to contain $5.85 \% NaCl$ and $9.50 \% MgCl _{2}$ be weight of solution. Calculate its normal boiling point assuming $80 \%$ ionisation for $NaCl$ and $50 \%$ ionisation of $MgCl _{2}\left[ K _{b}\left( H _{2} O \right)\right. =0.51\, kg\, mole^{ -1} K ]$.

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Solution:

$100\, gm \longrightarrow 5.85\, gm\, NaCl =.01$ mole $NaCl$
$100\, gm \longrightarrow 9.5\, gm\, MgCl { }_{2}=.1$ mole $MgCl _{2}$
wt. of water $=100-5.85-9.5=84.65\, gm$
$\Delta T _{ b }=(0.51) \frac{\frac{5.85}{58.5}(1+0.82)+\frac{9.5}{95}(1+0.5)}{84.651}$
$T _{ b } -100=(0.51)\left[\frac{0.18+0.15}{84.651}\right)(1000)$
$=\frac{(0.51)(0.33)(1000)}{84.651}$
$T _{ b } =100+1.988$
$T _{ b } =101.988^{\circ} C$