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K_(f) for water is 1.86^(@)C*m^(-1). What is the molality of a solutionn which freezes at -0.192^(@)C? Assuming no change in the solute by raising the temperature, at what temperature will the solution boil ? (K_(b) for H_(2)O=0.515^(@)C*m^(-1)) |
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Answer» SOLUTION :MOLALITY `=(DeltaT_(f))/(K_(f))` `=(0.192)/(1.86)=0.103m` Again `DeltaT_(B)=K_(b).m` …………(Eqn. 8) `=0.515xx0.103=0.0532^(@)C` Assuming the b.p. of pure water to be `100^(@)C` the b.p. of the solution will be `100.0532^(@)C` |
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