question_answer

Volume of \[\text{0}\text{.1}\,\text{M}\,{{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\] required to oxidize 35 mL of \[\text{0}\text{.5 M FeS}{{\text{O}}_{\text{4}}}\]solution is

A)  29.2 mL      

B)  17.5 mL

C)  175 mL      

D)  145 Ml

Correct Answer: A

Solution :

 \[C{{r}_{2}}O_{7}^{2-}+6F{{e}^{2+}}+{{14}^{+}}\xrightarrow{{}}6F{{e}^{3+}}\] \[+\,2C{{r}^{3+}}+7{{H}_{2}}O\] Hence, 1 mol of \[C{{r}_{2}}O_{7}^{2-}=6\]mole of \[\frac{{{M}_{1}}{{V}_{1}}}{1}=\frac{{{M}_{2}}{{V}_{2}}}{6}\] \[\frac{0.1\times {{V}_{1}}}{1}=\frac{0.5\times 35}{6}\] \[{{V}_{1}}=\frac{0.5\times 35}{6\times 0.1}\] \[{{V}_{1}}=29.2\,mL\]