A) 29.2 mL
B) 17.5 mL
C) 175 mL
D) 145 mL
Correct Answer: A
Solution :
\[C{{r}_{2}}O_{7}^{2-}+6F{{e}^{2+}}+14{{H}^{+}}\xrightarrow[{}]{{}}6F{{e}^{3+}}\] \[+2C{{r}^{3+}}+7{{H}_{2}}O\] Hence, 1 mol of\[C{{r}_{2}}O_{7}^{2-}=6\]mole of\[F{{e}^{2+}}\] \[\frac{{{M}_{1}}{{V}_{1}}}{1}=\frac{{{M}_{2}}{{V}_{2}}}{6}\] \[\frac{0.1\times {{V}_{1}}}{1}=\frac{0.5\times 35}{6}\] \[{{V}_{1}}=\frac{0.5\times 35}{6\times 0.1}\] \[{{V}_{1}}=29.2\,mL\]
You need to login to perform this action.
You will be redirected in
3 sec
