To find the concentration of the new solution, we need to go through the process step by step:
1. Calculate the moles of solute in the stock solution:
- Given mass of [tex]\((NH_4)_2SO_4\)[/tex]: [tex]\(66.05 \text{ g}\)[/tex]
- Molar mass of [tex]\((NH_4)_2SO_4\)[/tex]: [tex]\(132.1 \text{ g/mol}\)[/tex]
- Moles of solute [tex]\(\text{(n)} = \frac{\text{mass}}{\text{molar mass}} = \frac{66.05 \text{ g}}{132.1 \text{ g/mol}} = 0.5 \text{ mol}\)[/tex]
2. Calculate the molarity (M) of the stock solution:
- Volume of the stock solution: [tex]\(250 \text{ mL} = 0.250 \text{ L}\)[/tex]
- Molarity [tex]\(= \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.5 \text{ mol}}{0.250 \text{ L}} = 2.0 \text{ M}\)[/tex]
3. Use the dilution equation to find the concentration of the new solution:
- A 10.0 mL sample of the stock solution is taken and diluted to 50.0 mL.
- Initial volume [tex]\( (V_i) = 10.0 \text{ mL} = 0.010 \text{ L} \)[/tex]
- Final volume [tex]\( (V_f) = 50.0 \text{ mL} = 0.050 \text{ L} \)[/tex]
- Initial molarity [tex]\(M_i = 2.0 \text{ M} \)[/tex]
- Using the equation [tex]\(M_iV_i = M_fV_f\)[/tex]:
[tex]\[
2.0 \text{ M} \times 0.010 \text{ L} = M_f \times 0.050 \text{ L}
\][/tex]
- Solve for the new molarity [tex]\(M_f\)[/tex]:
[tex]\[
M_f = \frac{2.0 \text{ M} \times 0.010 \text{ L}}{0.050 \text{ L}} = 0.4 \text{ M}
\][/tex]
Therefore, the concentration of the new solution is [tex]\( 0.40 \text{ M} \)[/tex].
