Question

A solution of non volatile solute has boiling point elevation \(0.70\) K if \(K_b\) for the solvent is \(2.44 \text{K kg mol}^{-1}\). What is the molality of the solution?

• A. \(3.48\) m
• B. \(2.86\) m
• C. \(0.286\) m
• D. \(0.0186\) m

Hint:

Use delta Tb = Kb times molality.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The elevation in boiling point of a dilute solution of a non-volatile solute is proportional to its molality: \(\Delta T_b = K_b\, m\).

Step 2: Rearrange:
\[ m = \frac{\Delta T_b}{K_b} \]

Step 3: Substitute:
\[ m = \frac{0.70}{2.44} = 0.2869 \approx 0.286\ \text{m} \]

Step 4: Check the Other Options:
3.48 m is \(2.44/0.70\), the ratio inverted. 2.86 m is ten times too large, a decimal slip. 0.0186 m does not match the formula. Option (C) is correct.

Final Answer:
The molality is 0.286 m, option (C). \[ \boxed{\text{(C) } 0.286\ \text{m}} \]