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Empirical & Molecular Formula

Sea water is ...

Question

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

A

T_{b} = {51.15}^{\circ} C

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B

T_{b} = {78.6}^{\circ} C

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C

T_{b} = {102.3}^{\circ} C

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D

T_{b} = {157.2}^{\circ} C

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Solution

The correct option is C $T_{b} = {102.3}^{\circ} C$
The molar masses of NaCl and $M g C l_{2}$ are 58.5 g/mol and 95 g/mol respectively. 100 g of solution will contain 5.85 g NaCl and 9.5 g $M g C l_{2}$ and $100 - 5.85 - 9.5 = 84.65$ g water.

The number of moles of NaCl are $\frac{5.85 g}{58.5 g / m o l} = 0.1$

The number of moles of $M g C l_{2}$ are $\frac{9.5 g}{95 g / m o l} = 0.1$

80% ionisation of NaCl will give $0.1 \times 2 \times 0.8 + 0.1 \times 0.2 = 0.16 + 0.02 = 0.18$ moles of particles and 50% ionization of $M g C l_{2}$ will give $0.1 \times 0.5 \times 3 + 0.1 \times 0.5 = 0.15 + 0.5 = 0.2$ moles of particles.

Hence, the molarity of the solution $= \frac{0.18 + 0.2 m o l e s}{0.08465 k g} = 4.489 m$

The elevation in the boiling point of the solution is $Δ T_{b} = K_{b} \times m = 0.51 \times 4.489 = 2.3$

Hence, the boiling point of the solution is $100 + 2.3 = 102.3^{0} C$

Hence, the correct option is $C$

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