A) below \[0{}^\circ C\]
B) \[0{}^\circ C\]
C) \[1{}^\circ C\]
D) \[2{}^\circ C\]
Correct Answer: A
Solution :
\[Pb{{\left( N{{O}_{3}} \right)}_{2}}\] ionises to produce ions in solution so its freezing point will be below \[0{}^\circ C\].
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