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:
Answer & explanation
Correct answer: option 3
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.