Sea water is found to contain 5.85 % NaCl and 9.50% MgCl 2 by weight of solution. Calculate its normal boiling point assuming 80% ionisation for NaCl and 50% ionisation of MgCl 2 (K b (H 2 O) = 0.51 kgmol –1 K).
Text Solution
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( b ) Δ T b = K b .m.i
In 100 g of solution
moles of NaCl = 0.1 ( α = 0.8)
moles of MgCl 2 = 0.1 ( α = 0.5)
NaCl ⎯→ Na + + Cl –
i NaCl = 1 + (2 – 1) 0.8 = 1.8
Effective no. of moles of NaCl = 0.1 × 1.8 = 0.18
= 1 + (3–1) 0.5 = 2
Effective no. of moles of MgCl 2 = 0.1 ×2 = 0.2
Total no.of mole = 0.18 + 0.2 = 0.38
Δ T b = 
× 1000 × 0.51 = 3.8 × 0.51 = 2.28
So, T b = 100 + 2.28 = 102.28 ≈ 102.3
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