Question

In a titration experiment,$10\,mL$ of an $FeCl_2$ solution consumed $25 \,mL$ of a standard $K_2Cr_2O_7$ solution to reach the equivalent point. The standard $K_2Cr_2O_7$ solution is prepared by dissolving $1.225\,g$ of $K_2Cr_2O_7$ in $250 \,mL$ water. The concentration of the $FeCl_2$ solution is closest to [Given : molecular weight of$K_2Cr_2O_7 = 294\,g \,mol^{-1}]$

• A. 0.25 N
• B. 0.50 N
• C. 0.10 N
• D. 0.04 N

Answer 1

Answer: (A)

In titration at equivalence point, number of equivalents of both reactants must be same.
$\therefore N_{1}V_{1}=N_{2}V_{2} $
$ N$ (normality) of $FeCl_{2} $
$ =\frac{\left(N\times V\right) \,\text{of} \,K_{2}Cr_{2}O_{7}}{V \,\text{of}\, FeCl_{2}} $
$N \left(\text{normality}\right) \,\text{of}\, K_{2}Cr_{2}O_{7}$
$=\frac{ \text{Mass}\times100}{\text{Equivalent mass}\times \text{Volume of solution prepared}}$
Equivalent mass $=\frac{\text{Molar mass}}{n\, \text{factor}}$
$=\frac{294}{6}$
$ = 49\,g/$ equi.
$(K_2Cr_2O_7$ as oxidising agent is reduced to $2 \,Cr^{3+})$.
Oxidation state of $2$ Cr atoms change from + $6$ to + $3$.
Total change is $3 \times$2=6$,\left(\therefore n-\text{factor}=6\right)$
$\therefore N\left( \text{of} K_{2}Cr_{2}O_{7}\right)=\frac{1.225\times1000}{49\times250}=0.1N$
$\therefore N\left( \text{of} FeCl_{2}\right)=\frac{0,1\times25}{10}$
$=0.25\,N$