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Lake water is found to contain 9% $\mathrm{NaCl}$ , 3.8% $\mathrm{MgCl}_2$ and 5.21% $\mathrm{NaHCO}_3$ by weight of solution. Calculate its boiling point assuming 90% ionization of $\mathrm{NaCl}$ , 60% ionization of $\mathrm{MgCl}_2$ and 30% ionization of $\mathrm{NaHCO}_3$ kb for water = 0.52
Option: 1
$108.5^{\circ} \mathrm{C}$

Option: 2
$122.6^{\circ} \mathrm{C}$

Option: 3
$133.4^{\circ} \mathrm{C}$

Option: 4
$102.25^{\circ} \mathrm{C}$

Answers (1)

best_answer

$\mathrm{n_{NaCl}}=\frac{9}{58.5}=0.1$

${\mathrm{n_{MgCl_2}}}=\frac{3.8}{95}=0.04$

$\mathrm{n_{NaHCO_3}}=\frac{5.2}{84}=0.06 \\$

$\mathrm{ i_{NaCl} }=1+\alpha=1+0.9=1.9 \\$

$\mathrm{i_{MgCl_2}}=1+2 \alpha=1+0.6 \times 2=2.2$

$\mathrm{i_{Na_2 {HCO}_3}}=1+\alpha=1+0.3=1.3 \\$

$\text { Weight of solvent }=100-\{9+3.8+5.2\}$

$=100-18$

$=82 \mathrm{~g} \text {. } \\$

$\frac{1000 \times 0.2 \times(1.9 \times 0.1+2.2 \times 0.04+1.3 \times 0.06)}{82}$

$\mathrm{ \Delta T_b=\frac{1000 \times 0.52 \times 0.356}{82} \\ }$

$=2.25^{\circ} \mathrm{C} \\$

$\therefore \text { Boiling pt of solution }=100+2.25 \\$

$=102.25^{\circ} \mathrm{C} \\$

Posted by

Pankaj

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