To solve the problem step-by-step, let's follow these logical steps:
### Step 1: Calculate the moles of solute in the initial solution.
Given:
- Mass of solute (ammonium sulfate, [tex]\((NH_4)_2SO_4\)[/tex]): [tex]\( 66.05 \)[/tex] grams
- Molar mass of [tex]\((NH_4)_2SO_4\)[/tex]: [tex]\( 132.1 \)[/tex] g/mol
We can calculate the moles of [tex]\((NH_4)_2SO_4\)[/tex] using the formula:
[tex]\[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass}} \][/tex]
[tex]\[ \text{moles of solute} = \frac{66.05 \text{ g}}{132.1 \text{ g/mol}} \][/tex]
[tex]\[ \text{moles of solute} = 0.5 \text{ moles} \][/tex]
### Step 2: Calculate the molarity of the initial solution.
Given:
- Volume of the initial solution: [tex]\( 250 \)[/tex] mL (which is [tex]\( 0.250 \)[/tex] L)
- Moles of solute: [tex]\( 0.5 \)[/tex] moles
We use the formula for molarity (concentration, [tex]\(M\)[/tex]):
[tex]\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]
[tex]\[ M_{\text{initial}} = \frac{0.5 \text{ moles}}{0.250 \text{ L}} \][/tex]
[tex]\[ M_{\text{initial}} = 2.0 \text{ M} \][/tex]
### Step 3: Calculate the molarity of the final diluted solution using [tex]\(M_i V_i = M_f V_f\)[/tex].
Given:
- The initial sample volume ([tex]\(V_i\)[/tex]): [tex]\( 10.0 \)[/tex] mL (which is [tex]\( 0.010 \)[/tex] L)
- The final volume of the diluted solution ([tex]\(V_f\)[/tex]): [tex]\( 50.0 \)[/tex] mL (which is [tex]\( 0.050 \)[/tex] L)
- The molarity of the initial solution: [tex]\( 2.0 \)[/tex] M
Using the dilution formula:
[tex]\[ M_i V_i = M_f V_f \][/tex]
Solving for [tex]\( M_f \)[/tex] (final molarity):
[tex]\[ M_f = \frac{M_i V_i}{V_f} \][/tex]
[tex]\[ M_f = \frac{2.0 \text{ M} \times 0.010 \text{ L}}{0.050 \text{ L}} \][/tex]
[tex]\[ M_f = \frac{0.02 \text{ M \cdot L}}{0.050 \text{ L}} \][/tex]
[tex]\[ M_f = 0.4 \text{ M} \][/tex]
Thus, after diluting, the concentration of the new solution is [tex]\( 0.4 \text{ M} \)[/tex].
So, the correct answer is:
[tex]\[ 0.400 \text{ M} \][/tex]
