Question

Q.2. a) On dissolving 2.25 gm glucose (molecular weight = 180) in 25
gm water, the freezing point was lowered by 0.93 C. Calculate the molal
depression constant for water.​

Answer 1

Given info : On dissolving 2.25 g glucose (molecular weight = 180) in 25 g water, the freezing point was lowered by 0.93°C.

To find : The molal depression constant for water is ...

solution : mass of glucose = 2.25 g

molecular mass of glucose = 180 g/mol

so no of moles of glucose = 2.25/180 = 0.0125 mol

mass of solvent (water) = 25 g

now molality of solution ,m = no of moles of glucose/mass of water in kg

= 0.0125 mol/(25/1000 kg)

= 0.0125 × 1000/25

= 0.5

the depression in freezing point, ∆T_f = 0.93°C

using formula, ∆T_f = k_f × m

⇒0.93°C = k_f × 0.5 molal

⇒k_f = 1.86 °C kg/mol

Therefore the molal depression of water is 1.86 °C kg/mol or 1.86 k kg/mol.

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