Question

Dilution by mass is more accurate than dilution by volume. A stocksolution contains 1.044 g Fe/kg solution in 0.48 M HCl. How many $\mathbf{\mu gFe}\mathbf{/}\mathbf{g}$solution are in a solution made by mixing 2.145 g stock solution with 243.27 g 0.1 M HCl?

Short answer

The amount of$\mathrm{\mu gFe}/\mathrm{g}$is$9.125\mathrm{\mu gFe}/\mathrm{g}$

Step-by-step solution

Calculate the Mass:

$\mathrm{Amount}\mathrm{Fe}(\mathrm{\mu g}/\mathrm{g})=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Fe}}{\mathrm{Mass}\mathrm{of}\mathrm{solution}}.......\left(1\right)$

Equation for mass of solution,

Mass of solution = Mass of stock solution + Mass of HCL

Given,

Mass of stock solution =2.145 g

Mass of HCL=243.27 g

Mass of solution $=2.145\mathrm{gn}+243.27\mathrm{g}$

$\mathrm{Mass}\mathrm{of}\mathrm{solution}=245.415\mathrm{g}$

Mass of Fe in stock solution = (Mass of Fe per kg solution)(Mass of stock solution)

Mass of Fe in stock solution = $\left(\frac{1.044\mathrm{g}}{1\mathrm{kg}}\right)\left(\frac{1\mathrm{kg}}{{10}^3\mathrm{g}}\right)$

Mass of Fe in stock solution =$0.00223938\mathrm{g}$

Substitute in equation (1),

STEP:2 :  Calculate  the amount of iron (Fe)

$\mathrm{Amount}\mathrm{of}\mathrm{Fe}(\mathrm{\mu g}/\mathrm{g})=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Fe}}{\mathrm{Mass}\mathrm{of}\mathrm{solution}}$

role="math" localid="1663669292492" $$\begin{gathered} \mathrm{Amount}\mathrm{of}\mathrm{Fe}(\mathrm{\mu g}/\mathrm{g})=\left(\frac{0.00223938}{245.415\mathrm{g}}\right)\left(\frac{{10}^6}{1\mathrm{g}}\right) \\ \mathrm{Amount}\mathrm{of}\mathrm{Fe}(\mathrm{\mu g}/\mathrm{g})\approx 9.125\mathrm{\mu g}/\mathrm{g} \end{gathered}$$