A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ------- g mol\(^{-1}\) (Nearest integer).
Given: Water boils at 373 K, \(K_b\) for water = 0.52 K kg mol\(^{-1}\).
A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ------- g mol\(^{-1}\) (Nearest integer).
Given: Water boils at 373 K, \(K_b\) for water = 0.52 K kg mol\(^{-1}\).
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When solving boiling point elevation problems, always ensure the masses are converted into proper units (grams to kilograms) and the molality formula is applied correctly. Use the given values of Kb and ∆Tb to calculate the molar mass of the solute accurately.
Correct Answer: 100
Solution and Explanation
Step 1: Boiling Point Elevation
The elevation in boiling point (\(\Delta T_b\)) is given as:
\[\Delta T_b = T_b - T_b^\circ = 373.52 - 373 = 0.52\,\text{K}.\]
Step 2: Use the Boiling Point Elevation Formula
The boiling point elevation is related to molality (\(m\)) by:
\[\Delta T_b = K_b \cdot m.\]
Substitute \(m = \frac{\text{mass of solute (g)}}{\text{Molar Mass (g mol}^{-1}) \times \text{mass of solvent (kg)}}\):
\[0.52 = 0.52 \cdot \frac{2}{\text{Molar Mass} \times 20 \times 10^{-3}}.\]
Step 3: Solve for Molar Mass
Simplify:
\[1 = \frac{2}{\text{Molar Mass} \times 0.02}.\]
\[\text{Molar Mass} = \frac{2}{0.02} = 100 \, \text{g mol}^{-1}.\]
Conclusion: The molecular mass of the solute is \(100 \, \text{g mol}^{-1}\).
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