# How do we prepare a 100*mL of a 0.010*mol*L^-1 solution of Fe^(2+) in aqueous solution...?

Jun 6, 2017

If we use $\text{Mohr's salt}$, ${\left(N {H}_{4}\right)}_{2} F e {\left(S {O}_{4}\right)}_{2} \cdot 6 {H}_{2} O$ whose molar mass is $392.13 \cdot g \cdot m o {l}^{-} 1$. We need approx. $4 \cdot g$ to prepare such a solution.

#### Explanation:

A $0.10 \cdot N$ solution $\equiv$ $0.10 \cdot m o l \cdot {L}^{-} 1$ is required.

Which, given a volume of $100 \cdot m L$ constitutes a molar quantity of....

$\equiv 0.10 \cdot m o l \cdot {L}^{-} 1 \times 100 \times {10}^{-} 3 \cdot L = 0.010 \cdot m o l$

Which constitutes a mass of................

$0.010 \cdot m o l \times 392.13 \cdot g \cdot m o {l}^{-} 1 = 3.921 \cdot g$

Note that $\text{Mohr's salt}$ is the standard $F e \left(+ I I\right)$ salt used in the laboratory, as it is stable with respect to oxidation.