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.