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: Formula for elevation in boiling point.
$\Delta T_b = i \times K_b \times m$, where $i$ is the van't Hoff factor, $K_b$ is the ebullioscopic constant, and $m$ is the molality.
Step 2: Determine the van't Hoff factor.
NaCl completely dissociates: $NaCl \rightarrow Na^+ + Cl^-$, so $i = 2$.
Step 3: Calculate $\Delta T_b$.
$\Delta T_b = 2 \times 0.52 \times 1 = 1.04^\circ C$.
Step 4: Find boiling point of solution.
Boiling point $= 100 + 1.04 = 101.04^\circ C$.
\[ \boxed{101.04^\circ C} \]