Question

when 2.00 moles of a molecular compound was dissolved in 1000. g of water. what is the freezing point of the solution? (freezing point constant of water is 1.86 °c/m)
3.72 °c
-0.00372 °c
0.00372 °c
-3.72 °c

Explanation:

Step1: Calculate molality (m)

Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \). Mass of water is 1000 g = 1 kg, moles of solute = 2.00 mol. So \( m = \frac{2.00\ \text{mol}}{1.000\ \text{kg}} = 2.00\ m \).

Step2: Calculate freezing point depression (\(\Delta T_f\))

For molecular compounds, van't Hoff factor \( i = 1 \). Formula: \( \Delta T_f = i \times K_f \times m \). \( K_f = 1.86\ ^\circ\text{C}/m \), \( i = 1 \), \( m = 2.00\ m \). So \( \Delta T_f = 1 \times 1.86\ ^\circ\text{C}/m \times 2.00\ m = 3.72\ ^\circ\text{C} \).

Step3: Determine freezing point of solution

Freezing point of pure water is \( 0^\circ\text{C} \). Solution freezes at \( 0^\circ\text{C} - \Delta T_f = 0^\circ\text{C} - 3.72^\circ\text{C} = -3.72^\circ\text{C} \).

Answer:

-3.72 °C (corresponding to the option: -3.72 °C)