Two solutions containing 0.75 g of urea (molecular weight 60) and 1.5 g of compound A in 100 g water, freeze at the same temperature. The molecular weight of A is:

120

The correct answer is option 3. 120.

To determine the molecular weight of compound A, we can use the concept of freezing point depression in solutions:

The freezing point depression (\(\Delta T_f\)) is related to the molal concentration of the solute in the solvent:

\(\Delta T_f = K_f \cdot m \)

Where:

\( \Delta T_f \) is the freezing point depression,

\( K_f \) is the cryoscopic constant (specific to the solvent, here assumed to be water),

\( m \) is the molality of the solution (moles of solute per kilogram of solvent).

Given:

Solution 1: Contains 0.75 g of urea (molecular weight \( M_{\text{urea}} = 60 \)) in 100 g of water.

Solution 2: Contains 1.5 g of compound A in 100 g of water.

Since both solutions freeze at the same temperature, they have the same freezing point depression, \( \Delta T_f \).

For Solution 1 (Urea):

\( \text{moles of urea} = \frac{\text{mass of urea}}{\text{molecular weight of urea}} = \frac{0.75 \text{ g}}{60 \text{ g/mol}} = 0.0125 \text{ mol} \)

\(m_1 = \frac{\text{moles of urea}}{\text{mass of water in kg}} = \frac{0.0125 \text{ mol}}{0.1 \text{ kg}} = 0.125 \text{ mol/kg} \)

For Solution 2 (Compound A):

\(m_2 = \frac{\text{moles of A}}{\text{mass of water in kg}} = \frac{\frac{1.5 \text{ g}}{M_A \text{ g/mol}}}{0.1 \text{ kg}} = \frac{15}{M_A} \text{ mol/kg} \)

Since both solutions freeze at the same temperature, their molalities must be equal:
\(m_1 = m_2 \)

\( 0.125 = \frac{15}{M_A} \)

Solving for \( M_A \):

\(M_A = \frac{15}{0.125} = 120 \text{ g/mol} \)

Therefore, the molecular weight of compound A is \(120\) g/mol.