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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)` |
Answer» `"n"_("NaCl")=11.7/58.5=0.2, " n"_("MgCl"_(2))=9.5/9.5=0.1, " n"_("NaHCO"_(3))=8.4/8.4=0.1` `i_(NaCl)=1+alpha=1+0.9=1.9`, i_(MgCl_(2))=1+2alpha=1+0.7xx2=2.4`, 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` |
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