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Avogadro's Law & Gay Lussac's Law

Sea water is ...

Question

Sea water is found to contain 5.85 per NaCl and 9.50 per $M g C l_{2}$ by weight of solution. Calculate its normal boiling point assuming 80 per ionisation for NaCl and 50 per ionisation of $M g C l_{2} [M_{b} (H_{2} O) = 0.51 k g m o l^{- 1} K]$ .

T_{f} = - {0.73}^{\circ} C

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T_{f} = + {0.73}^{\circ} C

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T_{f} = - {0.87}^{\circ} C

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T_{f} = + {0.87}^{\circ} C

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Solution

The correct option is A $T_{f} = - {0.73}^{\circ} C$
$Δ T_{b} = T_{b (s o l u t i o n)} - T_{b (s o l v e n t)} = i K_{b} \times m$
$K_{b} = 0.51 k g m o l^{-} K$
moles of $N a C l = 5.85 g / 58.5 g m o l^{-} = 0.1 m o l$ dissolved in $100 - 5.85 g m$ water
moles of $M g C l_{2} = 9.5 g / 95 g m o l^{-} = 0.1 m o l$ dissolved in $100 - 9.5 g m$ water.
$\Rightarrow$ molality of $N a C l = m = 0.1 / 0.094 k g = 1.06 m$
Molality of $M g C l_{2} = 0.1 / 0.0905 = 1.11$
use boiling point elevation constant:
$Δ T = i K_{b} m$
$x = (0.5 + 0.8) (0.51) (1.06 + 1.11) = {1.41}^{o} C$

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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)

11.7 %

N a C l

9.5 %

M g C l_{2}

8.4 %

N a H C O_{3}

90 %

N a C l

70 %

M g C l_{2}

50 %

N a H C O_{3}

K_{b}

= 0.52

m o l^{- 1}

_{1}^{3} T

T^{+}

β

- {0.372}^{\circ} C

(T_{f}^{\circ} - T_{f})

T_{f} indicates freezing point of solution

T_{f}^{\circ} indicates freezing point of pure solvent

0.518

887

T_{f}

T_{b}

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