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Sea water is found to contain 5.85% NaCI...

Sea water is found to contain `5.85% NaCI` and`9.50% MgCI_(2)` by weight of solution. Calculate its normal boiling point assuming `80%` ionisation for `NaCI` and `50%` ionisation of `MgCI[K_(b)(H_(2)O) =0.51 kg mol^(-1)K]`

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The correct Answer is:
`T_(b)=102.3^(@)"C"`
`Delta"T"_("b")=[1+("n"_("NaCl")-1)alpha_("NaCl")xx("W"_("NaCl"))/("m"_("NaCl"))+(1+("n"_("MgCl"_(2))-1)alpha_("MgCl"_(2)))("W"_("MgCl"_(2)))/("m"_("MgCl"_(2)))]("K"_("b")xx1000)/("W"_("solvent"))`
`[(1+0.8)xx(5.85)/(5.85)+(1+2xx0.5)(9.52)/(119)]xx(0.51xx1000)/(84.64)=2.04`
`therefore"Boiling point of sea water=102.01"^(@)"C"`
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