Respuesta :
Answer: Therefore, the concentration of final solution is [tex]4.0\times 10^{-8}M[/tex]
Explanation:
According to the neutralization law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = [tex]2.3\times 10^{-5}M[/tex]
[tex]V_1[/tex] = volume of stock solution = 3.00 ml
[tex]M_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 25.00 ml
[tex]2.3\times 10^{-5}M\times 3.00=M_2\times 25.00[/tex]
[tex]M_2=0.28\times 10^{-5}M[/tex]
Therefore, the concentration of diluted solution is [tex]0.28\times 10^{-5}M[/tex]
2) On further dilution
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = [tex]0.28\times 10^{-5}M[/tex]
[tex]V_1[/tex] = volume of stock solution = 3.00 ml
[tex]M_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 25.00 ml
[tex]0.28\times 10^{-5}M\times 3.00=M_2\times 25.00[/tex]
[tex]M_2=0.034\times 10^{-5}M[/tex]
Therefore, the concentration of diluted solution is [tex]0.034\times 10^{-5}M[/tex]
3) On further dilution
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = [tex]0.034\times 10^{-5}M[/tex]
[tex]V_1[/tex] = volume of stock solution = 3.00 ml
[tex]M_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 25.00 ml
[tex]0.034\times 10^{-5}M\times 3.00=M_2\times 25.00[/tex]
[tex]M_2=4.0\times 10^{-8}M[/tex]
Therefore, the concentration of final solution is [tex]4.0\times 10^{-8}M[/tex]
Molarity means the moles are divides by the volume of the solution. The molarity of the final dilution [tex]4.0\times10^-^8\;M[/tex]
What is molarity?
Molarity is the measure of the concentration of any solute in per unit volume of the solution.
By neutralization law
[tex]\rm M_1V_1=M_2V_2[/tex]
1. First dilution
M1 is molarity of stock solution = [tex]2.3 \times10^-^5 M[/tex]
V1 is volume of stock solution = 3.00 ml
V2 is volume of diluted solution = 25.00 ml
M2 molarity of diluted solution = ?
[tex]\rm (2.3 \times10^-^5)\times 3.00=M_2 \times 25.00[/tex]
[tex]M_2 = 0.28\times 10^-^5[/tex]
2. second dilution
M1 is molarity of stock solution = [tex]0.28\times 10^-^5[/tex]
V1 is volume of stock solution = 3.00 ml
V2 is volume of diluted solution = 25.00 ml
M2 molarity of diluted solution = ?
[tex]\rm (0.28\times 10^-^5)\times 3.00=M_2 \times 25.00\\\\M_2 = 0.034\times 10^-5[/tex]
3. Third dilution
M1 is molarity of stock solution = [tex]0.034\times 10^-5[/tex]
V1 is volume of stock solution = 3.00 ml
V2 is volume of diluted solution = 25.00 ml
M2 molarity of diluted solution = ?
[tex]\rm (0.034\times 10^-5)\times 3.00=M_2 \times 25.00\\\\M_2 = 4.0\times 10^-^8[/tex]
The molarity of the final dilution = [tex]4.0\times 10^-^8[/tex]
Learn more about molarity, here:
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