Question

The molal freezing point constant for water is 1.86. If 342 g of cane sugar is dissolved in 1000 g of water, the solution will freeze at ______.

पर्याय

• 1.86°C

• −1.86°C

• −3.92°C

• 3.92°C

Answer 1

The molal freezing point constant for water is 1.86. If 342 g of cane sugar is dissolved in 1000 g of water, the solution will freeze at −1.86°C.

Explanation:

Given: Mass of solute (cane sugar) = 342 g

Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 342 g/mol

Mass of water (solvent) = 1000 g = 1 kg

Molality (m) = `"Mass of solute"/("Molar mass" xx "Mass of solvent in kg")`

= `342/(342 xx 1)`

Molality (m) = 1 mol/kg

Freezing point depression constant (K_f) = 1.86 K kg/mol

Freezing point of pure water = 0°C

ΔT_f = K_f m

= 1.86 × 1 

= 1.86°C

Freezing point = 0°C − 1.86°C

= −1.86°C