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:
Answer & explanation
Correct answer: option 4
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.