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]`

Options

A. `T_(b)=101.9^(@)c`
B. `T_(b)=104.9^(@)C`
C. `T_(b)=108.5^(@)C`
D. `T_(b)=110.3^(@)C`

Answer

A

Explanation

`DeltaT_(b)=K_(b).m.i`
In `100gm` of solution
moles of `NaCl=0.1`
`(alpha=0.8)`
moles of `MgCl_(2)=0.1`
`(alpha=0.5)`
`NaClrarrNa^(+)+Cl^(-)`
`i_(NaCl)=1+(2-1)0.8=1.8`
Effective no. of moles of `MgCl_(2)=0.1xx1.8=0.18`
`i_(MgCl_(2))=1+(3-1)0.5=2`
Effective no. of moles of `MgCl_(2)=0.1xx2=0.2`
Total no. of mole `=0.18+0.2=0.38`
`DeltaT_(b)=(0.38)/(100)xx100xx0.51=3.8xx0.51=1.938`
So, `T_(b)=100+1.938=101.938`