The freezing point depression of 645 g of aqueous solution of ethylene glycol \(\left( C _2 H _6 O _2\right)\) is 2.25 K . Find weight of ethylene glycol in the solution. \(\left[K_{r}=1.86 K kg mol^{-1} ; H=1, C=12, O=16 amu\right]\)

1. Molar Mass Calculation:
The molecular formula of ethylene glycol is \( C_2H_6O_2 \).
Molar Mass \( (M_2) = (2 \times 12) + (6 \times 1) + (2 \times 16) \)
\( M_2 = 24 + 6 + 32 = 62 \text{ g mol}^{-1} \)
2. Setup:
Let the mass of ethylene glycol (solute) be \( w_2 \) grams.
Total mass of solution = \( 645 \text{ g} \).
Mass of water (solvent), \( w_1 = (645 - w_2) \text{ g} \).
Given: \( \Delta T_f = 2.25 \text{ K} \) and \( K_f = 1.86 \text{ K kg mol}^{-1} \).
3. Calculation:
Using the freezing point depression formula:
\( \Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1} \)

Substituting the values:
\( 2.25 = \frac{1.86 \times w_2 \times 1000}{62 \times (645 - w_2)} \)
\( 2.25 = \frac{30 \times w_2}{645 - w_2} \)
\( 2.25(645 - w_2) = 30w_2 \)
\( 1451.25 - 2.25w_2 = 30w_2 \)
\( 1451.25 = 32.25w_2 \)
\( w_2 = \frac{1451.25}{32.25} = 45 \text{ g} \)
Correct Option: 45 g