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Mole fraction of `C_(3)H_(5)(OH)_(3)` in a solution of `36g` of water and `46g` of glycerine is:A. `0.46`B. `0.36`C. `0.20`D. `0.40` |
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Answer» Correct Answer - C Mole of `H_(2)O=(36)/(18)=2` Mole of glycerine `=(46)/(92)=0.5` total mole `=2+0.5=2.5` Mole fractions of glycerine `=(n_(1))/(n_(1)+n_(2))=(0.5)/(2.5)rArr X_(0)=0.2` |
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