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The mole fraction of a solute in bezene solvent is 0.2. The molality of the solution will be:A. 14B. 3.2C. 1.4D. 2 |
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Answer» Correct Answer - b `X_(B)=n_(B)/(n_(B)+n_(A))` `0.2=n_(B)/(n_(b)+1000//78)` `1/0.2=(n_(B)+12.82)/n_(B)` `5=1+(12.82)/n_(B)n_(B)=(12.82)/4=3.2` (Mass of solvent is 1000 g) |
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