Question

$15\mathrm{mL}$ of an aqueous solution of ${\mathrm{Fe}}^{+2}$ in the acidic medium completely reacted with $20\mathrm{mL}$ of $0.03\mathrm{M}$ aqueous $C{r}_2{O_7}^{-2}$. The molarity of ${\mathrm{Fe}}^{+2}$ solution is $\mathrm{\_\_\_\_\_\_}\mathrm{x}{10}^{-2}\mathrm{M}$.(Round off to the Nearest Integer).

Answer 1

Step 1: Given data:

Volume of ${\mathrm{Fe}}^{+2}\left(\mathrm{aq}\right)$ in acidic medium is ${\mathrm{V}}_1=15\mathrm{mL}$

Volume of $C{r}_2{O_7}^{-2}\left(\mathrm{aq}\right)$ is ${\mathrm{V}}_2=20\mathrm{mL}$

Molarity of ${\mathrm{Fe}}^{+2}$ is ${\mathrm{M}}_1=\mathrm{\_\_\_\_\_\_}\mathrm{x}{10}^{-2}\mathrm{M}$

Molarity of $C{r}_2{O_7}^{-2}$ is ${\mathrm{M}}_2=0.03\mathrm{M}$

Step 2: Balanced chemical equation is

$C{r}_2{O_7}^{-2}+14{\mathrm{H}}^{+}+6{\mathrm{Fe}}^{+2}\to 2{\mathrm{Cr}}^{+3}+7{\mathrm{H}}_2\mathrm{O}+6{\mathrm{Fe}}^{+2}$

From the above equation. ${\mathrm{n}}_1=1\mathrm{mol}$and ${\mathrm{n}}_2=6\mathrm{mol}$

Step 3: Find the molarity of ${\mathbf{Fe}}^{\mathbf{+}\mathbf{2}}\mathbf{}$:

By using the law of equivalence; ${\mathit{M}}_{\mathit{e}\mathit{q}}ofF{e}^{+2}=M_{eq}ofC{r}_2{O_7}^{-2}$

$$\begin{gathered} \Rightarrow {\left({\mathrm{M}}_1\times {\mathrm{V}}_1\times {\mathrm{n}}_1\right)}_{F{e}^{+2}}={\left({\mathrm{M}}_2\times {\mathrm{V}}_2\times {\mathrm{n}}_2\right)}_{C{r}_2{O_7}^{-2}} \\ \Rightarrow \left(\mathrm{x}\times {10}^{-2}\right)M\times 15\mathrm{mL}\times 1\mathrm{mol}=0.03\mathrm{M}\times 20\mathrm{mL}\times 6\mathrm{mol} \\ \Rightarrow \mathrm{x}=\frac{0.03\times 20\times 6}{{10}^{-2}\times 15\times 1} \\ \Rightarrow \mathrm{x}=24 \end{gathered}$$

Hence, the molarity of ${\mathbf{Fe}}^{\mathbf{+}\mathbf{2}}\mathbf{}$is $\underline{\underline{\mathbf{24}}}\mathbf{}\mathbf{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}\mathbf{M}$.