 # In a titrant experiment, 10 ml of $FeC{l_2}$ solution consumed 25ml of standard ${K_2}C{r_2}{O_7}$ solution to reach the equivalence point. This standard ${K_2}C{r_2}{O_7}$ solution is prepared by dissolving 1.225g of ${K_2}C{r_2}{O_7}$ in 250ml water. The concentration of the $FeC{l_2}$ solution is closest to:[Given: molecular weight of ${K_2}C{r_2}{O_7}$ = 204g $mo{l^{ - 1}}$]A.0.25NB.0.50NC.0.10ND.0.04N Verified
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Hint: We need to remember that when two chemicals react with each other the formation of the product is going to depend on the energy required to convert chemical reactants into products. If the energy to convert reactants into products is just too high, then the reaction isn’t going to happen.

As we know that the ferric chloride may be a common compound of iron and chlorine during which iron possesses +3 oxidation state. From the question we’ve,
Given: V = 250 ml, $V_1$ = 10 ml, $V_2$ = 25 ml, $N_1$ =? m1 = 1.225g and M = 294g/ mol
We know that $N = M \times {n_f}$
When $FeC{l_2}$ reacts with ${K_2}C{r_2}{O_7}$ its $F{e^{2 + }}$ ion gets converted into $F{e^{3 + }}$ ion.
From the above equation,
No. of equivalent of $FeC{l_2}$ = no. of equivalent of ${K_2}C{r_2}{O_7}$
Therefore, $N_1V_1 = N_2V_2$
Substituting the values of $V_1$ and $V_$2 we get,
$\Rightarrow {N_1} \times 10 = M \times {n_f} \times 25$
${N_1} \times 10 = \dfrac{n}{{V(L)}} \times {n_f} \times 25$
On substituting the known values we get,
${N_1} \times 10 = \dfrac{{1.225/294}}{{250 \times {{10}^{ - 3}}}} \times 6 \times 25$
On simplification we get,
${N_1} = 0,25N$
$FeC{l_2} + {K_2}C{r_2}{O_7} \to F{e^{ + 3}} + C{r^{ + 3}}$

Note:
-We have to remember that the IUPAC name of ferric chloride is Iron (III) chloride or iron trichloride. Apart from ferric chloride it’s several common names like molysite and Flores martis. It possesses various colours like as in its anhydrous form it appears black-green or purple while in its hydrous form it appears yellow solid.
-The formation of the products in a chemical reaction is going to depend on ionization energy of the chemicals. If the reaction needs the highest amount of ionization energy we’ve to use a catalyst to complete the reaction.