Question

1.
By dissolving 13.6 g of a substance in 20g of water, the freezing point is decreased by 3.7°c Then,the
molecular mass of the substance is (Molal depression constant of water = 1.863 K kgmol-1​

Answer 1

Answer:

Depression in Freezing Point-

∆Tf = I kf m

I = 1

kf = 1.86

∆Tf = 3.7

m = molality

m = 2

molality = Mole/weight of Solvent in Kg

weight of Solvent in Kg- 20 × 10^-3Kg

mole = Weight /Molecular weight

weight - 13.6 gram(Given)

Molecular Weight = 340

Any doubt Comment

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Answer 2

Explanation:

$\sf Mw_2 = \dfrac{1000K_f×W_2}{W_1×∆T}$

$\sf K_f =1.863 \: K \: kg \: mol^{-1}$

$\sf Mw_2 = \dfrac{1000×1.863×13.6}{20×3.7}$

$\sf = 342.39$