Question

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

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

Hint:

Always check if the solute is an electrolyte. If it is, never forget to multiply by \( i \).
Common values: \( NaCl \) (\( i=2 \)), \( MgCl_2 \) (\( i=3 \)), Glucose (\( i=1 \)).
For boiling point, the answer must be greater than \( 100^\circ C \).

Answer 1

The Correct Option is B

Step 1: Note that NaCl gives two particles per formula unit.
Since $NaCl \rightarrow Na^+ + Cl^-$ splits completely into two ions, the van't Hoff factor here is $i = 2$.
Step 2: Apply the elevation formula with this factor included.
\[ \Delta T_b = iK_bm = 2 \times 0.52 \times 1 = 1.04^\circ C \]
Step 3: Add this rise to the normal boiling point of water.
\[ T_b = 100 + 1.04 = 101.04^\circ C \]
Step 4: State the final boiling point.
The salty solution boils at a higher temperature than pure water because of the extra dissolved particles. \[ \boxed{101.04^\circ C} \]