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A stock solution is made by dissolving [tex]66.05 \, \text{g}[/tex] of [tex](\text{NH}_4)_2\text{SO}_4[/tex] in enough water to make [tex]250 \, \text{mL}[/tex] of solution. A [tex]10.0 \, \text{mL}[/tex] sample of this solution is then diluted to [tex]50.0 \, \text{mL}[/tex]. Given that the molar mass of [tex](\text{NH}_4)_2\text{SO}_4[/tex] is [tex]132.1 \, \text{g/mol}[/tex], what is the concentration of the new solution?

Use [tex]M_i V_i = M_f V_f[/tex] and molarity [tex]= \frac{\text{moles of solute}}{\text{liters of solution}}[/tex].

A. [tex]0.400 \, M[/tex]
B. [tex]1.60 \, M[/tex]
C. [tex]5.00 \, M[/tex]
D. [tex]10.0 \, M[/tex]



Answer :

GinnyAnswer
To find the concentration of the new solution, we can follow these steps:

1. Calculate the moles of solute in the initial solution:

Given:
- The mass of solute (ammonium sulfate, [tex]\((NH_4)_2SO_4\)[/tex]) is [tex]\( 66.05 \)[/tex] grams.
- The molar mass of [tex]\((NH_4)_2SO_4\)[/tex] is [tex]\( 132.1 \)[/tex] [tex]\(\text{g/mol}\)[/tex].

The number of moles is calculated using the formula:
[tex]\[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass}} \][/tex]
[tex]\[ \text{moles of solute} = \frac{66.05 \text{ g}}{132.1 \text{ g/mol}} = 0.500 \text{ mol} \][/tex]

2. Convert the initial volume of the solution from mL to L:

Given:
- The initial volume of the solution is [tex]\( 250 \)[/tex] mL.

To convert mL to L:
[tex]\[ \text{Volume in liters} = \frac{\text{Volume in mL}}{1000} \][/tex]
[tex]\[ \text{Volume in liters} = \frac{250 \text{ mL}}{1000} = 0.250 \text{ L} \][/tex]

3. Calculate the molarity of the initial solution:

Molarity (M) is given by:
[tex]\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]
[tex]\[ M = \frac{0.500 \text{ mol}}{0.250 \text{ L}} = 2.00 \text{ M} \][/tex]

4. Determine the volume of the sample taken from the initial solution and its molarity:

Given:
- The volume of the sample taken is [tex]\( 10.0 \)[/tex] mL.
- The volume of the diluted solution is [tex]\( 50.0 \)[/tex] mL.

First, convert the sample volume and the diluted solution volume to liters:
[tex]\[ \text{Volume of the sample in liters} = \frac{10.0 \text{ mL}}{1000} = 0.010 \text{ L} \][/tex]
[tex]\[ \text{Volume of the diluted solution in liters} = \frac{50.0 \text{ mL}}{1000} = 0.050 \text{ L} \][/tex]

5. Use the dilution formula [tex]\( M_i V_i = M_1 V_f \)[/tex] to calculate the molarity of the new solution:

Where:
- [tex]\( M_i \)[/tex] = Initial molarity of the solution = [tex]\( 2.00 \text{ M} \)[/tex]
- [tex]\( V_i \)[/tex] = Initial volume of the sample = [tex]\( 0.010 \text{ L} \)[/tex]
- [tex]\( M_1 \)[/tex] = Final molarity of the new solution = ?
- [tex]\( V_f \)[/tex] = Final volume of the new (diluted) solution = [tex]\( 0.050 \text{ L} \)[/tex]

Rearrange the dilution formula to solve for [tex]\( M_1 \)[/tex]:
[tex]\[ M_1 = \frac{M_i V_i}{V_f} \][/tex]
Substitute the given values:
[tex]\[ M_1 = \frac{2.00 \text{ M} \times 0.010 \text{ L}}{0.050 \text{ L}} = 0.40 \text{ M} \][/tex]

Therefore, the concentration of the new solution is [tex]\( 0.400 \text{ M} \)[/tex]. The correct answer is [tex]\( 0.400 \text{ M} \)[/tex].

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