Answer :
To find the concentration of the new solution, let's go through the problem step-by-step.
Step 1: Calculate the number of moles of [tex]\((NH_4)_2SO_4\)[/tex] in the stock solution.
Given:
- Mass of [tex]\((NH_4)_2SO_4 = 66.05 \, \text{g}\)[/tex]
- Molar mass of [tex]\((NH_4)_2SO_4 = 132.1 \, \text{g/mol}\)[/tex]
The number of moles of [tex]\((NH_4)_2SO_4\)[/tex] can be calculated using the formula:
[tex]\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{moles of } (NH_4)_2SO_4 = \frac{66.05 \, \text{g}}{132.1 \, \text{g/mol}} \approx 0.5 \, \text{moles} \][/tex]
Step 2: Calculate the initial concentration of the stock solution.
Given:
- Volume of the solution = 250 mL = 0.250 L
The concentration (molarity) of the stock solution is:
[tex]\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]
[tex]\[ \text{Initial concentration} = \frac{0.5 \, \text{moles}}{0.250 \, \text{L}} = 2.0 \, \text{M} \][/tex]
Step 3: Evaluate the effect of diluting the stock solution.
A 10.0 mL sample of the 2.0 M stock solution is diluted to 50.0 mL.
Using the dilution formula [tex]\( M_i V_i = M_f V_f \)[/tex]:
Given:
- [tex]\(M_i = 2.0 \, \text{M}\)[/tex]
- [tex]\(V_i = 10.0 \, \text{mL} = 0.010 \, \text{L}\)[/tex]
- [tex]\(V_f = 50.0 \, \text{mL} = 0.050 \, \text{L}\)[/tex]
We need to find the final concentration [tex]\(M_f\)[/tex]:
[tex]\[ M_i V_i = M_f V_f \][/tex]
[tex]\[ 2.0 \, \text{M} \times 0.010 \, \text{L} = M_f \times 0.050 \, \text{L} \][/tex]
[tex]\[ 0.020 = M_f \times 0.050 \][/tex]
[tex]\[ M_f = \frac{0.020}{0.050} = 0.4 \, \text{M} \][/tex]
Conclusion:
The concentration of the new solution after dilution is [tex]\( 0.4 \, \text{M} \)[/tex]. Therefore, the correct answer is:
[tex]\[ 0.400 \, \text{M} \][/tex]
Step 1: Calculate the number of moles of [tex]\((NH_4)_2SO_4\)[/tex] in the stock solution.
Given:
- Mass of [tex]\((NH_4)_2SO_4 = 66.05 \, \text{g}\)[/tex]
- Molar mass of [tex]\((NH_4)_2SO_4 = 132.1 \, \text{g/mol}\)[/tex]
The number of moles of [tex]\((NH_4)_2SO_4\)[/tex] can be calculated using the formula:
[tex]\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{moles of } (NH_4)_2SO_4 = \frac{66.05 \, \text{g}}{132.1 \, \text{g/mol}} \approx 0.5 \, \text{moles} \][/tex]
Step 2: Calculate the initial concentration of the stock solution.
Given:
- Volume of the solution = 250 mL = 0.250 L
The concentration (molarity) of the stock solution is:
[tex]\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]
[tex]\[ \text{Initial concentration} = \frac{0.5 \, \text{moles}}{0.250 \, \text{L}} = 2.0 \, \text{M} \][/tex]
Step 3: Evaluate the effect of diluting the stock solution.
A 10.0 mL sample of the 2.0 M stock solution is diluted to 50.0 mL.
Using the dilution formula [tex]\( M_i V_i = M_f V_f \)[/tex]:
Given:
- [tex]\(M_i = 2.0 \, \text{M}\)[/tex]
- [tex]\(V_i = 10.0 \, \text{mL} = 0.010 \, \text{L}\)[/tex]
- [tex]\(V_f = 50.0 \, \text{mL} = 0.050 \, \text{L}\)[/tex]
We need to find the final concentration [tex]\(M_f\)[/tex]:
[tex]\[ M_i V_i = M_f V_f \][/tex]
[tex]\[ 2.0 \, \text{M} \times 0.010 \, \text{L} = M_f \times 0.050 \, \text{L} \][/tex]
[tex]\[ 0.020 = M_f \times 0.050 \][/tex]
[tex]\[ M_f = \frac{0.020}{0.050} = 0.4 \, \text{M} \][/tex]
Conclusion:
The concentration of the new solution after dilution is [tex]\( 0.4 \, \text{M} \)[/tex]. Therefore, the correct answer is:
[tex]\[ 0.400 \, \text{M} \][/tex]
