Question

A sample of seawater was found to contain $9.5\%w/w$ of $MgCl_{2}$ and $5.85\%w/w$ of $NaCl$ . Assuming complete ionization of salts, what will be the boiling point of sample. ( $K_{b}$ for water $=0.52$ , at. wt. of $Mg=24$ , at. wt. of $Na=23$ ,
at. wt. of $Cl=35.5$ ) :-

• A. $101.5^\circ C$
• B. $103.07^\circ C$
• C. $100.52^\circ C$
• D. $105.08^\circ C$

Answer 1

Answer: (B)

Consider $100gm$ of solution
the mass of $MgCl_{2}=9.5gm$ & mass of $NaCl=5.85gm$
Hence mass of solvent $=100-\left(9 . 5 + 5 . 85\right)=84.65$
for $MgCl_{2},i_{1}=3$ & for $NaCl,i_{2}=2$
$\left(ΔT\right)_{b}=\frac{0 . 52 \times \left(3 \times \frac{9 . 5}{95} + 2 \times \frac{5 . 85}{58 . 5}\right)}{84 . 65 \times \left(10\right)^{- 3}}$
$\left(ΔT\right)_{b}=\frac{0 . 52 \times \left(3 \times 0 . 1 + 2 \times 0 . 1\right)}{84 . 65 \times \left(10\right)^{- 3}}$
$=\frac{0 . 52 \times 0 . 5}{84 . 65 \times 10^{- 3}}=3.07^\circ C$
$\therefore $ Boiling point of sea water sample $=100+3.07$ $=103.07^\circ C$