By dissolving 5 g substance in 50 g of water, the decrease in freezing point is 1.2°C. The molal depression constant is 1.85° kg mol^–1. The molecular weight of a substance is:

154.2

The correct answer is option 4. 154.2.

To determine the molecular weight of the substance, we can use the formula for depression in freezing point, which is given by:

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

where:

\(\Delta T_f\) = decrease in freezing point

\(i\) = van 't Hoff factor (which is 1 for non-electrolytes)

\(K_f\) = molal depression constant

\(m\) = molality of the solution

First, let's solve for the molality \(m\) using the given data:

\(\Delta T_f = 1.2^\circ \text{C}\)

\(K_f = 1.85 \, \text{°C kg mol}^{-1}\)

Rearranging the formula to solve for molality \(m\):

\(m = \frac{\Delta T_f}{K_f \cdot i}\)

Since \(i = 1\) (assuming the substance does not dissociate), the formula simplifies to:

\(m = \frac{1.2}{1.85} \approx 0.648 \, \text{mol/kg}\)

Molality \(m\) is defined as the number of moles of solute per kilogram of solvent. Given that 50 g of water is the solvent (which is 0.050 kg), the number of moles \(n\) can be calculated by:

\(n = m \times \text{kg of solvent} = 0.648 \times 0.050 \approx 0.0324 \, \text{moles}\)

The molecular weight (M) of the substance is given by:

\(\text{Molecular weight} = \frac{\text{mass of substance}}{\text{number of moles}}\)

Given that 5 g of the substance is used:

\(\text{Molecular weight} = \frac{5 \, \text{g}}{0.0324 \, \text{moles}} \approx 154.2 \, \text{g/mol}\)

Conclusion: The molecular weight of the substance is approximately 154.2 g/mol.