KCET · Chemistry · Some Basic Concepts of Chemistry
Which of the following aqueous solution has highest freezing point?
- A \( 0.1 \) molal \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \)
- B \( 0.1 \) molal \( \mathrm{BaCl}_{2} \)
- C \( 0.1 \) molal \( \mathrm{AlCl}_{3} \)
- D \( 0.1 \) molal \( \mathrm{NH}_{4} C l \)
Answer & Solution
Correct Answer
(D) \( 0.1 \) molal \( \mathrm{NH}_{4} C l \)
Step-by-step Solution
Detailed explanation
Freezing point depression is directly proportional to the no. of dissolved particles, \( \Delta T_{f}=i k_{f} m \). But freezing point is inversely proportional to no. of dissolved particles
$\begin{array}{|l|c|} \hline \text{ Compounds } & \text{ Number of lons or van't Hoff Factor (i) } \\ \hline \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \rightarrow 2 \mathrm{Al}^{3+}+3 \mathrm{SO}_{4}{ }^{2-} & 5 \\ \hline \mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} & 3 \\ \hline \mathrm{AlCl}_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} & 3\\ \hline \mathrm{NH}_{4} \mathrm{Cl} \rightarrow \mathrm{NH}_{4}+\mathrm{Cl}^{-} & 1 \\ \hline \end{array}
Therefore, it can be concluded that \( \mathrm{NH}_{4} \mathrm{Cl} \) has a highest freezing point as \( \mathrm{i}=2 \).
$\begin{array}{|l|c|} \hline \text{ Compounds } & \text{ Number of lons or van't Hoff Factor (i) } \\ \hline \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \rightarrow 2 \mathrm{Al}^{3+}+3 \mathrm{SO}_{4}{ }^{2-} & 5 \\ \hline \mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} & 3 \\ \hline \mathrm{AlCl}_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} & 3\\ \hline \mathrm{NH}_{4} \mathrm{Cl} \rightarrow \mathrm{NH}_{4}+\mathrm{Cl}^{-} & 1 \\ \hline \end{array}
Therefore, it can be concluded that \( \mathrm{NH}_{4} \mathrm{Cl} \) has a highest freezing point as \( \mathrm{i}=2 \).
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