River water is found to contain `11.7%` `NaCl`, `9.5% MgCl_(2)`, and `8.4%`. `NaHCO_(3)`, by weight of solution. Calculate its normal boiling point assuming `90%` ionization of `NaCl`, `70%` ionization of `MgCl_(2)` and `50%` ionization of `NaHCO_(3)(K_(b)` for water `=0.52)`

Explanation

`"n"_("NaCl")=11.7/58.5=0.2, " n"_("MgCl"_(2))=9.5/9.5=0.1, " n"_("NaHCO"_(3))=8.4/84=0.1`
`i_(NaCl)=1+alpha=1+0.9=1.9`,
`i_(MgCl_(2))=1+2alpha=1+0.7xx2=2.4`,
`i_(NaHCO_(3))=1+2alpha=1+0.5xx2=2.0`
Weight of solvent `=100-(11.7+9.5+8.4)=70.4g`
`Delta T_(b)=((i_(NaCl)xxn_(NaCl)+i_(MgCl_(2))xxn_(MgCl_(2))+i_(NaHCO_(3))xxn_(NaHCO_(3)))xxK_(b)xx1000)/("Weight of solvent")`
`=((1.9xx0.2+2.4xx0.1+2xx0.1)xx0.52xx1000)/70.4=5.94^(@)C`
`therefore` Boiling point of solution `=100+5.94=105.95^(@)C`