Given: Volume of Mohr's salt solution ($V_1$) = $25.0\text{ mL}$ Normality of $K_2Cr_2O_7$ solution ($N_2$) = $0.1\text{ N}$ Volume of $K_2Cr_2O_7$ solution ($V_2$) = $21.63\text{ mL}$ Using the titration formula $N_1 V_1 = N_2 V_2$: $$ \begin{aligned} N_1 \times 25 &= 0.1 \times 21.63 \\ N_1 &= \frac{2.163}{25} = 0.08652\text{ N} \end{aligned} $$ Amount of Iron (Fe) in $25\text{ mL}$ solution: $$ \begin{aligned} \text{Mass of Fe} &= N_1 \times V_1 \times 55.85 \times 10^{-3} \\ &= 0.08652 \times 25 \times 55.85 \times 10^{-3} \\ &= 0.1207\text{ g} \end{aligned} $$ (Note: Mohr's salt is $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$)