A) \[A{{B}_{2}}\]
B) \[1.0M\text{ }HCl\]
C) \[100mL\]
D) \[0.05M\]
Correct Answer: A
Solution :
\[Al,\] i for \[NaOH/air,\Delta \] i for \[N{{a}_{2}}C{{O}_{3}}/air,\Delta \] i for \[N{{H}_{4}}Cl,\Delta \] i for \[Al\] i for \[N{{a}_{2}}C{{O}_{3}}/air,\Delta \] Thus, depression in freezing point wil be in order \[{{V}_{C}}=3b\ \,equal\] Hence, freezing point will be in order \[=3\times 0.056=0.168d{{m}^{3}}mo{{l}^{-1}}\]
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