A stock solution is made by dissolving 66.05 g of [tex]\((NH_4)_2SO_4\)[/tex] in enough water to make 250 mL of solution. A 10.0 mL sample of this solution is then diluted to 50.0 mL. Given that the molar mass of [tex]\((NH_4)_2SO_4\)[/tex] is 132.1 g/mol, what is the concentration of the new solution?

Use [tex]\(M_iV_i = M_fV_f\)[/tex] and [tex]\(\text{molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}\)[/tex].

A. 0.400 M
B. 1.60 M
C. 5.00 M
D. 10.0 M



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]