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Standard VIII

Chemistry

Boiling

Sea water is ...

Question

Sea water is found to contains 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 $M g C l_{2} [K_{b} (H_{2} O) = 0.5 k g m o l^{- 1} K]$

T_{b} = 101.3 K

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T_{b} = 102 . 24^{\circ} C

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T_{b} = 112 K

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T_{b} = 110^{\circ} C

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Solution

The correct option is B $T_{b} = 102 . 24^{\circ} C$
100 $g$ sea water contains 5.85 $g N a C l (0.1 m o l e)$ and 9.5 $g M g C l_{2} (0.1 m o l e)$
$∴ effective molality = \frac{0.1 (1.8) + 0.12 (2)}{(100 - 5.85 - 9.5)} \times 100$
= 4.48
$∴ Δ T_{b} = m_{e f f} \times k_{b} = {2.24}^{\circ} C$
$T_{b} = {102.24}^{\circ} C$

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Sea water is found to contains 5.85% NaCl and 9.50% MgCl2 by weight of solution. Calculate its normal boiling point assuming 80% ionisation for NaCl and 50% ionisation of MgCl2[Kb(H2O)=0.5 kg mol−1 K]

The correct option is B Tb=102.24∘ C100 g sea water contains 5.85 g NaCl (0.1 mole) and 9.5 g MgCl2 (0.1 mole) ∴ effective molality=0.1(1.8)+0.12(2)(100−5.85−9.5)×100 = 4.48 ∴ ΔTb=meff×kb=2.24∘C Tb=102.24∘C

Sea water is found to contains 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 $M g C l_{2} [K_{b} (H_{2} O) = 0.5 k g m o l^{- 1} K]$