InterviewSolution
Saved Bookmarks
| 1. |
The addition of `0.643 g` of a compound to `50 mL` of benzene (density 0.879 g `mL^(-1)`) lowers the freezing point from `5.51` to `5.03^(@)C`. If `K_(f)` for benzene is `5.12`, calculate the molecular weight of the compound. |
|
Answer» Correct Answer - `Mw=156.06 g mol^(-1)` According to depression in freezing point method `Mw_(2)=(1000xxK_(f))/(DeltaT_(f)) xx W_(2)/W_(1)` Given that `K_(f) rarr` molar depression constant of benzene `=5-12 K kg "mol"^(-1)` `W_(2)` =Weight of compound =0.643 g `W_(1)` =Weight of benzene =volume`xx` Density `=50 xx 0.879` `=43.95 g` `DeltaT_(f) rarr` Depression in freezing point =`5.51-5.03=0.48^(@)C` Therefore, on substitution, `Mw_(2)=(1000xx5.12xx0.643)/(0.48xx43.95)=156.06` |
|