Question

An acid of general molecular formula HA is \(25\%\) dissociated during the study of boiling point of its solution. If the calculated elevation of boiling point of the solution is \(1.62\,K\), find the observed value of elevation of boiling point of the solution.

• A. \(2.025\,K\)
• B. \(2.430\,K\)
• C. \(2.632\,K\)
• D. \(2.835\,K\)

Hint:

For dissociation: \[ i=1+\alpha(n-1) \] where \(n\) is the number of particles formed. For \[ HA \rightarrow H^+ + A^- \] \[ n=2 \] so, \[ i=1+\alpha \]

Answer 1

The Correct Option is A

Step 1: Identify what changes.
A colligative property depends on the number of particles. Dissociation makes more particles, so the observed elevation is larger than the calculated one by the van't Hoff factor $i$.

Step 2: Write the dissociation.
The acid breaks as $HA \rightleftharpoons H^+ + A^-$, giving $2$ particles from $1$.

Step 3: Formula for i.
For a particle splitting into $2$ ions, \[ i = 1 + \alpha, \] where $\alpha$ is the degree of dissociation.

Step 4: Put in the numbers.
Here $\alpha = 0.25$, so \[ i = 1 + 0.25 = 1.25. \]
Step 5: Scale the elevation.
Observed value $= i \times$ calculated value $= 1.25 \times 1.62$.

Step 6: Compute.
\[ \Delta T_b(\text{observed}) = 1.25 \times 1.62 = 2.025\,K. \]
\[ \boxed{2.025\,K} \]