Question

A solution of glucose (molar mass 180 g mol 1) has been prepared by dissolving 36 g of glucose in 500 g of water. The freezing point of the solution obtained will be (Kf of water = 1.86 K kg mol-¹)​

Answer 1

Answer:

Given mass of glucose = 36 g

Molar mass of glucose = 180 g

Number of moles of glucose = Given mass / Molar mass

⇒ n = 36 g / 180 g

⇒ n = 0.2 moles

Also it is given that the mass of the solvent (water) is 500 g, which is 0.5 kg.

Calculating the molality of the glucose solution we get:

$\boxed{ \textbf{Molality (m)} = \dfrac{\textbf{No. of moles of solute}}{\textbf{Mass of solvent (in Kg)}} \bf{\times 100}}$

Substituting the values we get:

⇒ Molality (m) = (0.2 / 0.5) × 100

⇒ Molality (m) = 0.4 × 100 = 40 mol/kg

The formula to find depression in freezing point is:

⇒ ΔT = K(f) × Molality

⇒ ΔT = 1.86 K. kg/mol × 40 mol/kg

⇒ ΔT = 74.4 K

Now the new freezing point of the solution obtained is:

⇒ New Freezing Point = 273 K - ΔT

⇒ New Freezing Point = 273 K - 74.4 K

⇒ New Freezing Point = 198.6 K

Hence the new freezing point of the solution is 198.6 K (or) -74.4° C.