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Calculate the bolling point of one mole aqueous solution (density 1.06 g cm^(-3)) of KBr. (Given : K_(b), for H_(2)O =0.52K kg mol^(-1), Atomic mass : K =39 , Br = 80 ] |
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Answer» Solution :One molar aqueous solution BOILING POINT = 7 d = 1.06 g/ml assuming VOLUME 1 L of KBr. Weight of KBr = 1.06 g `DeltaK_(b)=0.52` K kg /mol . `DeltaT_(b)=iK_(b)` m weight of solvent 1 kg `DeltaT_(b) = T_(2)-T_(1)` where `T_(1)= 373 K` `DeltaT_(b) = iK_(b) xx(W_(B)xx1000)/(M_(B)W_(A))` `DeltaT_(b)=2xx0.52 xx(1.06xx1000)/(119xx1)` `DeltaT_(b)=(1.1024xx1000)/(119)` =9.26 `T_(2)-373=9.26` `=382.26 K` . |
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