Question:
0.01 M solution of KCl and CaCl\(_2\) are prepared in water. The freezing point of KCl is found to be \(-2^{\circ}C\). What is the freezing point of CaCl\(_2\) (assumed to be completely ionised)?
0.01 M solution of KCl and CaCl\(_2\) are prepared in water. The freezing point of KCl is found to be \(-2^{\circ}C\). What is the freezing point of CaCl\(_2\) (assumed to be completely ionised)?
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\(\Delta T_f = i K_f m\). Van't Hoff factor \(i\) = number of ions per formula unit.
Updated On: Apr 23, 2026
- \(-3^{\circ}C\)
- \(+3^{\circ}C\)
- \(-2^{\circ}C\)
- \(-4^{\circ}C\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Freezing point depression \(\Delta T_f = i \cdot K_f \cdot m\). For same molality, \(\Delta T_f \propto i\).
Step 2: Detailed Explanation:
KCl dissociates as KCl \(\rightarrow\) K\(^+\) + Cl\(^-\), \(i = 2\). Given \(\Delta T_f(\text{KCl}) = 2^{\circ}C\).
CaCl\(_2\) dissociates as CaCl\(_2\) \(\rightarrow\) Ca\(^{2+}\) + 2Cl\(^-\), \(i = 3\).
\(\frac{\Delta T_f(\text{CaCl}_2)}{\Delta T_f(\text{KCl})} = \frac{3}{2} \Rightarrow \Delta T_f(\text{CaCl}_2) = \frac{3}{2} \times 2 = 3^{\circ}C\).
Freezing point = \(0 - 3 = -3^{\circ}C\).
Step 3: Final Answer:
Thus, freezing point of CaCl\(_2\) = \(-3^{\circ}C\).
Freezing point depression \(\Delta T_f = i \cdot K_f \cdot m\). For same molality, \(\Delta T_f \propto i\).
Step 2: Detailed Explanation:
KCl dissociates as KCl \(\rightarrow\) K\(^+\) + Cl\(^-\), \(i = 2\). Given \(\Delta T_f(\text{KCl}) = 2^{\circ}C\).
CaCl\(_2\) dissociates as CaCl\(_2\) \(\rightarrow\) Ca\(^{2+}\) + 2Cl\(^-\), \(i = 3\).
\(\frac{\Delta T_f(\text{CaCl}_2)}{\Delta T_f(\text{KCl})} = \frac{3}{2} \Rightarrow \Delta T_f(\text{CaCl}_2) = \frac{3}{2} \times 2 = 3^{\circ}C\).
Freezing point = \(0 - 3 = -3^{\circ}C\).
Step 3: Final Answer:
Thus, freezing point of CaCl\(_2\) = \(-3^{\circ}C\).
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