You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relative Lowering of Vapour Pressure

River water i...

Question

River water is found to contain $11.7 %$ $N a C l$ , $9.5 %$ $M g C l_{2}$ , and $8.4 %$ . $N a H C O_{3}$ by weight of solution. Calculate its normal boiling point assuming $90 %$ ionization of $N a C l$ , $70 %$ ionization of $M g C l_{2}$ and $50 %$ ionization of $N a H C O_{3}$ ( $K_{b}$ for water $= 0.52$ K $m o l^{- 1}$ kg).

Open in App

Solution

$^{n} N a C l = \frac{11.7}{58.5} = 0.2$
$^{n} M g C l_{2} = \frac{9.5}{9.5} = 0.1$
$^{n} N a H C O_{3} = \frac{8.4}{84} = 0.1$
$^{i} N a C l = 1 + α = 1 + 0.9 = 1.9$
$^{i} M g C l_{2} = 1 + 2 α = 1 + 0.7 \times 2 = 2.4$
$^{i} N a H C O_{3} = 1 + 2 α = 1 + 0.5 \times 2 = 2.0$
Weight of the solvent = $100 - (11.7 = 9.5 + 8.4)$
= $70.4 g$
$Δ T_{b}$ = $\frac{(^{i} N a C l \times^{n} N a C l +^{i} m g c l_{2} +^{i} N a H C O_{3} \times^{n} N a H C O_{3}) \times K_{b} \times l o g}{w e i g h t o f s o l v e n t}$
$= \frac{(1.9 \times 0.2 + 2.4 \times 0.1 + 2 \times 0.1) \times 0.52 \times 1000}{70.4}$
$= {5.94}^{o} C$
Boiling point of solution = $100 + 5.94$
= ${105.95}^{o} C$

Suggest Corrections

Similar questions

5.85 %

9.50 %

M g C l_{2}

80 %

50 %

M g C l_{2}

(K_{b} (H_{2} O) = 0.51 k g m o l^{- 1} K)

M g C l_{2}

M g C l_{2} [M_{b} (H_{2} O) = 0.51 k g m o l^{- 1} K]

N a_{2} S O_{4} (M = 142 g m o l^{- 1})

N a_{2} S O_{4}

K_{b}

m o l^{- 1}

M g S O_{4} (M = 120 g m o l^{- 1})

M g S O_{4}

K_{b}

0.52 K k g m o l^{- 1}]

8.2 % N a_{3} P O_{4}

12 % M g S O_{4}

%

N a_{3} P O_{4}

M g S O_{4}

50

60

^{o} C

[k_{b} (H_{2} O) = 0.50 {K kg mol}^{- 1}

Join BYJU'S Learning Program