Sea water is found to contain 5.85% NaCl and 9.50% MgCl 2 by weight of solution. Calculate its normal boiling point (in °C) assuming 80% ionisation for NaCl and 50% ionisation of MgCl 2 [ K b (

Question

Sea water is found to contain 5.85% NaCl and 9.50% MgCl 2 ​ by weight of solution. Calculate its normal boiling point (in °C) assuming 80% ionisation for NaCl and 50% ionisation of MgCl 2 ​ [ K b ​ ( H 2 ​ O ) = 0.51 kgmol − 1 K ].

TestHub · Accepted Answer

Take 100 g of seawater:

NaCl = 5.85 g, MgCl₂ = 9.50 g, Water = 84.65 g = 0.08465 kg

For NaCl:

$m = \frac{5.85/58.5}{0.08465} = 1.18$

$i = 1 + (2 - 1) (0.8) = 1.8$

For MgCl₂:

$m = \frac{9.5/95}{0.08465} = 1.18$

$i = 1 + (3 - 1) (0.5) = 2$

Total effective molality:

$m_{e ff} = 1.18 (1.8) + 1.18 (2) = 4.484$

$Δ T_{b} = K_{b} m_{e ff} = 0.51 (4.484) = 2.29 K$

$T_{b} = 100 + 2.29 = 102.29℃$

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