Question

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 1

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