A) 25
B) 50
C) 60
D) 75
Correct Answer: D
Solution :
[d] \[C{{r}_{2}}O_{7}^{-2}+6F{{e}^{+2}}(n=1)(Moh{r}'s\,salt)\] \[+14{{H}^{\text{+}}}\xrightarrow{{}}2C{{r}^{+3}}+6F{{e}^{+3}}+7{{H}_{2}}O\] Equivalent of \[F{{e}^{+2}}\]= moles of Mohr's salt = equivalent of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] \[=500\times {{10}^{-3}}\times 6\times 1=3\] Hence mole percent of Mohr's salt \[=\frac{3}{4}\times 100=75\]You need to login to perform this action.
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