Question

find the freezing point of a solution containing 0.520 gram glucose dissolved in 80.2 gram of water

Answer 1

Molecular mass of glucose

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



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Answer 2

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