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

Concept:

• Boiling point elevation is a colligative property that depends on the number of solute particles in the solution.

• For electrolytic solutes that dissociate, we must account for the van't Hoff factor (\( i \)).

• The elevation in boiling point (\( \Delta T_{b} \)) is given by the formula: \[ \Delta T_{b} = i \times K_{b} \times m \]

• The final boiling point of the solution (\( T_{b} \)) is: \[ T_{b} = T_{b}^\circ + \Delta T_{b} \] where \( T_{b}^\circ \) is the boiling point of pure water (\( 100^\circ C \)).
Step 1: Determine the van't Hoff factor (\( i \))
The problem states that \( NaCl \) is completely dissociated.
\( NaCl \) dissociates as: \[ NaCl \rightarrow Na^{+} + Cl^{-} \] Since 1 mole of \( NaCl \) produces 2 moles of particles, \( i = 2 \).

Step 2: Identify the given values
Molality (\( m \)) = \( 1 \, molal \)
Ebullioscopic constant (\( K_{b} \)) = \( 0.52 \, K \, kg \, mol^{-1} \)
Boiling point of pure water (\( T_{b}^\circ \)) = \( 100^\circ C \) (or \( 373.15 \, K \))

Step 3: Calculate the elevation in boiling point (\( \Delta T_{b} \))
Using the formula: \[ \Delta T_{b} = 2 \times 0.52 \times 1 \] \[ \Delta T_{b} = 1.04 \, K \] (Note: A change of \( 1.04 \, K \) is identical to a change of \( 1.04^\circ C \)).

Step 4: Calculate the boiling point of the solution
\[ T_{b} = 100^\circ C + 1.04^\circ C \] \[ T_{b} = 101.04^\circ C \]