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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]`

Text Solution

Verified by Experts

The correct Answer is:
`T_(b) = 102.3^(@)C`
`T'_(b) - 100^(@)C = 0.51 xx ((1.8 xx (5.85)/(58.5)+2xx(9.5)/(95)))/(((100-5.85-9.5)/(1000)))`
`T_(b) = 100 +0.51 xx (0.38)/(84.65//1000)`
`= 100^(@)C +2.29 = 102.29^(@)C`
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