A) \[{{K}_{2}}S{{O}_{4}}\]
B) \[NaCl\]
C) Urea
D) Glucose
Correct Answer: A
Solution :
Key Idea: Depression in freeing point is a colligative properly. It depends on no. of particles. More the number of particles, more will be depression in freezing point. [a] \[{{K}_{2}}S{{O}_{4}}\xrightarrow{{}}2{{K}^{+}}+SO_{4}^{2-}\] It gives 3 particles [b] \[NaCl\xrightarrow{{}}N{{a}^{+}}+C{{l}^{-}}\] It gives 2 particles [c] Urea \[\xrightarrow{{}}\] No dissociation. [d] Glucose \[\xrightarrow{{}}\] No dissociation. \[\because \] \[{{K}_{2}}S{{O}_{4}}\] produces maximum number of particles \[\therefore \] \[{{K}_{2}}S{{O}_{4}}\] has maximum depression in freezing point.
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