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Mole fraction of `A` in `H_(2)O` is `0.2` The molality of `A` in `H_(2)O` is :A. `13.8`B. `15.5`C. `14.5`D. `16.8` |
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Answer» Correct Answer - A `x_(Lambda)=0.2` `x_(H_(2))O = 1-0.2=0.8` `"wt of" H_(2)O = 0.8xx18 = 14.4 g` Molality `=("moles of solute")/("wt. of solvent" (H_(2)O)"in kg")` `= (2xx1000)/(14.4)=13.8` |
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