Question

What is boiling point of solution containing \(7.45\ \text{g}\) of \(\text{KCl}\) in \(500\ \text{g}\) of water, assuming \(80\%\) dissociation of \(\text{KCl}\)? (molar mass of \(\text{KCl}=74.5\ \text{g/mole}\), \(k_{b}\) for water \(=0.52\ \text{K/molal}\))

• A. \(99.8128\ ^{\circ }\text{C}\)
• B. \(100.1872\ ^{\circ }\text{C}\)
• C. \(100.0936\ ^{\circ }\text{C}\)
• D. \(100.3744\ ^{\circ }\text{C}\)

Hint:

Find the van't Hoff factor from the degree of dissociation, then use delta Tb = i Kb m.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For an electrolyte, \(\Delta T_{b}=i\,K_{b}\,m\), where \(i=1+\alpha(n-1)\).

Step 2: Molality
Moles of KCl \(=\dfrac{7.45}{74.5}=0.1\).
Molality \(=\dfrac{0.1}{0.5}=0.2\ \text{m}\).

Step 3: Van't Hoff factor
KCl gives 2 ions, so \(i=1+0.8(2-1)=1.8\).

Step 4: Calculation
\(\Delta T_{b}=1.8\times 0.52\times 0.2=0.1872\ \text{K}\)
Boiling point \(=100+0.1872=100.1872\ ^{\circ}\text{C}\)

Final Answer:
The solution boils at 100.1872 degrees Celsius. \[ \boxed{\text{(B) }100.1872\ ^{\circ}\text{C}} \]