Saved Bookmarks
| 1. |
find the freezing point of a solution containing 0.520 gram of glucose dissolved in 80.2 gram of water Kf is equals to 1.86 kg/mole |
|
Answer» C6H12O6=72+12+96C6H12O6=72+12+96 ⇒180gmol−1⇒180gmol−1 ΔTf=Kf×WB×1000MB×WAΔTf=Kf×WB×1000MB×WA ⇒1.86×0.520×1000180×80.20⇒1.86×0.520×1000180×80.20 ⇒0.067 Brainlest PLZ |
|