The freezing point of an aqueous solution containing 23 g of $C_2H_5OH$ in 1000 g of water, is

(Given $K_f$ for water = $1.86\, K\, kg\, mol^{-1}$)

-0.93 °C

The correct answer is Option (2) → -0.93 °C

First, calculate the molality of the solution.

• Molar mass of C₂H₅OH = 46 g mol^-1
• Moles of ethanol = $\frac{23}{46}$ = 0.5 mol
• Mass of water = 1000 g = 1 kg

$\text{Molality } (m) = \frac{0.5}{1} = 0.5 \, \text{m}$

Now apply the freezing point depression formula:

$\Delta T_f = K_f \times m = 1.86 \times 0.5 = 0.93 \, \text{K}$

Since freezing point of pure water is $0^\circ\text{C}$,

$\text{Freezing point of solution} = 0 - 0.93 = -0.93^\circ\text{C}$