Question

18g of glucose is dissolved in 250g of water .The freezing point of the solution will be (of of water =1.86k/kg mol¹)

Answer 1

Freezing point is 272.4 K.

Explanation:

Weight of water w1 = 250 g

Weight of glucose w2 = 18 g

Molar mass M2 = 180 g/mol

Kf of water = 1.86 KKg/mol

∆Tf = Kf * w2 * 1000 / M2 * w1

    = 1.86 * 18 * 1000 / 180 * 250

   =  0.744

∆Tf = T⁰f - Tf

    = 273.15 - 0.744

  ∆Tf   = 272.404

Freezing point is 272.4 K.

Answer 2

Given:

The mass of glucose, w = 18 gm

The mass of water, W = 250 gm

Kf = 1.86 K/kg

To Find:

The freezing point of the glucose solution.

Calculation:

- We know that:

ΔTf = Kf × m

⇒ (T° - Tf) = Kf × (w × 1000) / (M.wt × W)

⇒ (0 - Tf) = 1.86 × (18 × 1000) / (180 × 250)

⇒ -Tf = 1.86 × 4 / 10

⇒ Tf = - 0.744

- So, the freezing point of the glucose solution is - 0.744°C