Question

calculate the Temperature at which of solution containing 54 gm of glucose C6H12O6 in 250 gm of water will freeze ( Kf = 1.86).....please solve this urgent please

Answer 1

molality of the glucose =no of mole of glucose in 1 kg of water
no of mole of glucose = weight of the glucose/molecular weight of the glucose = 54/180 = 0.3

delta Tf=Kfm
delta Tf= (1.86*0.3)/0.250 =2.232

temperature at which it will freeze=273 -2.232 K =270.768 K

Answer 2

Hey ⚛
✴here is ur answer✴
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Molecular mass of glucose

MB=72+12+96MB=72+12+96

=180gmol−1=180gmol−1

ΔTf=Kf×WB×1000MB×wBΔTf=Kf×WB×1000MB×wB

⇒1.86×54×1000180×250⇒1.86×54×1000180×250

⇒2.23⇒2.23

Freezing point of solution =0-2.23=−2.23∘C


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I hope it helps
#Prem✌✡❤