Question

The boiling point of one molal NaCl solution, assuming NaCl to be completely dissociated in water, is: ($K_b = 0.52$ K kg mol$^{-1}$)

• A. 100.52$^\circ$C
• B. 101.04$^\circ$C
• C. 100.04$^\circ$C
• D. 101.52$^\circ$C

Hint:

NaCl gives $i = 2$. $\Delta T_b = i K_b m = 2 \times 0.52 \times 1 = 1.04^\circ C$.

Answer 1

The Correct Option is B

Step 1: Concept
Elevation in boiling point is a colligative property: $\Delta T_b = i \times K_b \times m$.

Step 2: Analysis
NaCl completely dissociates: $NaCl \rightarrow Na^+ + Cl^-$, giving van't Hoff factor $i = 2$. Molality $m = 1$ mol kg$^{-1}$, $K_b = 0.52$ K kg mol$^{-1}$.

Step 3: Calculation
$\Delta T_b = i \times K_b \times m = 2 \times 0.52 \times 1 = 1.04^\circ C$. Boiling point of solution $= 100 + 1.04 = 101.04^\circ C$.

Final Answer: (B)