# River water is found to contain 11.7% NaCl, 9.5% $MgCl_2$, and 8.4%. $NaHCO_3$ by weight of solution. Calculate its normal boiling point assuming 90% ionization of NaCl,70% ionization of $MgCl_2$ and 50% ionization of $NaHCO_3$ ($K_b$ for water = 0.52 k kg $mol^{−1}$)

$n_{NaCl}=\large\frac{11.7}{58.5}$$=0.2 n_{MgCl_2}=\large\frac{9.5}{95}$$=0.1$
$\large\frac{11.7}{58.5}=\frac{8.4}{84}=$$0.1$
$i_{Nacl}=1+\alpha=1+0.9=1.9$
$i_{MgCl_2}=1+2\alpha=1+0.7\times 2=2.4$
$i_{NaHCO_3}=1+2\alpha=1+0.5\times 2=2.0$
$\Delta T_b=\large\frac{(1.9\times 0.2+2.4\times 0.1+2\times 0.1)\times 0.52\times 1000}{70.4}$
$\Rightarrow 6.0657^{\large\circ}$C
Boiling point of solution = 100 + 5.94 = $106.057^{\large\circ}$C
Hence (A) is the correct answer.