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The elevation in boiling point of a solution 13.44 g of `CuCI_(2)` in 1 kg of water using the information (Molecular mass or `CuCI_(2)=13.4 "and " K_(b)=0.52" K molal "^(-01)` will beA. `0.16`B. `0.05`C. `0.1`D. `0.2` |
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Answer» Correct Answer - a `{:((a),CuCI_(2),hArr,Cu^(2+),+,2CI^(-)),("Initial",CuCI_(2),,0,,0),("molar conc",,,,,1),("Eqm.molar",,,,,),("conc",(1-alpha),,alpha,, 2alpha):}` `i=(1-alpha)+alpha+2alpha=1+2alpha` Assuming 100% dissociation `(alpha=1)` `i-1+2xx1=3` `DeltaT_(b)=ixxK_(b)xxm` `=(3xx(0.52"K kg mol"^(-1))xx(13.44g))/((134.4 "g mol"^(-1))xx(1 kg))` `=0.156 K ~~0.16 K` |
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