A) \[NaCl>N{{a}_{2}}S{{O}_{4}}>N{{a}_{3}}P{{O}_{4}}\]
B) \[NaCl<N{{a}_{2}}S{{O}_{4}}<N{{a}_{3}}P{{O}_{4}}\]
C) \[NaCl>N{{a}_{2}}S{{O}_{4}}\,\,\approx \,\,N{{a}_{3}}P{{O}_{4}}\]
D) \[NaCl<N{{a}_{2}}S{{O}_{4}}=N{{a}_{3}}P{{O}_{4}}\]
Correct Answer: B
Solution :
Colligative property in decreasing order \[N{{a}_{3}}P{{O}_{4}}>N{{a}_{2}}S{{O}_{4}}>NaCl\] \[N{{a}_{3}}P{{O}_{4}}\to 3N{{a}^{+}}+PO_{4}^{3-}=4\] \[N{{a}_{2}}S{{O}_{4}}\to 2N{{a}^{+}}+SO_{4}^{2-}=3\] \[NaCl\to N{{a}^{+}}+C{{l}^{-}}=2\]
You need to login to perform this action.
You will be redirected in
3 sec
