We have a 0.145 M solution of iron(II) bromide (M_FeBr2 = 215.65 g.mol^-1).

How much of this solution do we need to get 4.23 g of solute?

Question

We have a 0.145 M solution of iron(II) bromide (M_FeBr2 = 215.65 g.mol^-1).

How much of this solution do we need to get 4.23 g of solute?

| General Chemistry 2

Accepted Answer

Iron(II) bromide: FeBr₂

n_FeBr2 = M_FeBr2 x V_FeBr2 and n_FeBr2 = $\frac{m_{FeBr 2}}{M_{FeBr 2}}$

⇒ M_FeBr2 x V_FeBr2= $\frac{m_{FeBr 2}}{M_{FeBr 2}}$

⇒ V_FeBr2(in L) = $\frac{m_{FeBr 2} (in g)}{M_{FeBr 2} (in g . {mol}^{- 1})}$ x M_FeBr2 (in mol.L^-1)

⇒ V_FeBr2(in L) = $\frac{4.23}{215.65 \times 0.145}$ = 1.35 x 10^-1 L

Midterm 1 - Question 5 | General Chemistry 2