Question

(i) $\underset{aq.suspension}{BaC{O}_3}+N{a}_{2}Cr{O}_4+AcOH\to \underset{ppt}{M}$
(ii) $\underset{aq.suspension}{BaC{O}_3}+N{a}_{2}Cr{O}_4+HCN\to \underset{ppt.}{N}$
Which is the correct option regarding the given reactions?

• A. M and N are the same and characterized as $BaCr{O}_4$
• B. M is $BaCr{O}_4$ and N stands for no precipitate
• C. M is $BaCr{O}_4$ and N is $BaC{O}_3$
• D. M stands for no precipitate and N is $BaCr{O}_4$

Answer 1

Answer: (A)

M and N are the same and characterized as $BaCr{O}_4$

(i) $\underset{aq.suspension}{BaC{O}_3}+N{a}_{2}Cr{O}_4+AcOH\to \underset{ppt}{B}aCr{O}_4+N{a}_{2}C{O}_3$

(ii) $\underset{aq.suspension}{BaC{O}_3}+N{a}_{2}Cr{O}_4+HCN\to \underset{ppt.}{B}aCr{O}_4+N{a}_{2}C{O}_3$

So M and N are $BaCr{O}_4$.

Hence option A is correct.