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The elevation in boiling point, when `13.44 g` of freshly prepared `CuCI_(2)` are added to one kilogram of water, is [Some useful data, `K_(b) (H_(2)O) = 0.52 kg K mol^(-1), "mol.wt of" CuCI_(2) = 134.4 gm]`A. `0.05`B. `0.10`C. `0.16`D. `0.20` |
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Answer» Correct Answer - C `x(CuCl_(2))=(13.44)/(134.4)=0.1` `CuCl_(2)(aq)hArrCu^(2+)(aq)+2Cl^(-)(aq)` `i = 3` `DeltaT_(b)=iK_(b)xxm` `=(3xx0.52xx0.1)/(1)=0.156~~0.16` |
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