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Calculate the boiling point of solution ...

Calculate the boiling point of solution when 4g of `MgSO_4 ("M =120 g mol"^(-1))` was dissolved in 100 g of water assuming `MgSO_4` undergoes complete ionisation. `(K_b" for water=0.52 K kg mol"^(-1))`

Text Solution

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Given, weight of solute, `W_(2)=4g`
weight of solvent, `W_(1)=100g`
i=2 (as `MgSO_4` dissociates completely into 2 ions)
`MgSO_4 Leftrightarrow Mg^(2+) +SO_(4)^(2-)`
`K_(b)=0.52" K kg mol"^(-1), M_(2)=120"g mol"^(-1)`
The elevation in boilling point of solution
`triangleT_(b)=(i xx K_(b) xx W_(2) xx 1000)/(M_(2) xx W_(1))`
where i=van.t Hoff factor
`W_(2)` =weight of solute
`W_(1)`= weight of solvent
`M_(2)`= molar mass of solute
`T_(b)^@`= boiling point of pure solvent
`=(2 xx 0.52 xx 4 xx 1000)/(120 xx 100)=0.346K`
Boiling point of pure water is `100^@C` or 373 K
`T_b=triangleT_(b)+T_(b)^@`
`T_(b)=373+0.346=373.346K`
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