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If 30g of a solute of molecular weight 154 is dissolved in 250g of benzene, what will be the boiling point of the resulting solution under atmospheric pressure . The molal boiling point elevation constant for benzene is 2.61^(@)C.m^(-1) and the boiling point of pure benzene is 80.1^(@)C |
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Answer» Solution :Suppose the elevation in B.p. is `DeltaT_(b)` Molality `=(DeltaT_(b))/(K_(b))=(DeltaT_(b))/(2.61)` …………(Eqn . 8) Moles of solute `=(30)/(154)` `:.` molality `=(30)/(154)xx(1000)/(250)` moles / `1000g` of solvent Thus, `(DeltaT_(b))/(2.61)=(30)/(154)xx(1000)/(250)` `DeltaT_(b)=2^(@)` Thus the b.p. of the resulting solution is `80.1+2=82.1^(@)C` |
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