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]`

Answer

`102.29^(@)C`

Explanation

`100g` solution contain `=5.85g NaCl`
`=0.1 "mol" NaCl`
`100g` solution contains `=9.50g MgCl_(2)=0.1 "mol" MgCl_(2)`
Hence, weight of solvent `(H_(2)O)=100-(5.85+9.50)=84.65g`
`NaClhArrNa^(+)+Cl^(-)`
`i=1+(y-1)x=(1+x)=1+0.8=1.8 y=2`
Here, `y` is the number of ions per moles of solute and `x` is the degree of ionisation. Hence, number of moles of `NaCl` from `0.1` mole due to ionisation `=1.8xx0.1=0.18` mol.
`MgCl_(2)` ionises `50%` MgCl_(2)hArrMg^(2+)+2Cl^(-)`
`therefore i=1+(y-1)x=1+2x=1+2xx0.5=2 y=3`
Hence, number of moles of `MgCl_(2)` in solution `=0.18+0.20`
or `(n_(1)+n_(2))i=0.38`
Elevation in b.p. `(DeltaT_(b))=(1000K_(b)(n_(1)+n_(2))i)/(w_(2))`
`=(1000xx0.51xx0.38)/(84.65)=2.29^(@)`.
Hence, b.p. of solution `=100+2.29^(@)`
`=102.29^(@)C`