Respuesta :
Answer:
The correct answer is 1.33 x 10⁻⁵ M
Explanation:
The concentration of the stock solution is: C= 1.33 M
In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:
C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M
The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
= 1.33 x 10⁻⁵ M
The final concentration of NaOH is 1.33 x 10⁻⁵ M this can be calculated by using the law of dilution.
What information do we have?
The concentration of the stock solution, C= 1.33 M
Law of dilution:
In the first dilution,
the student added 1 ml of stock solution for 9 ml of water. The alll- out volume of the arrangement is 1 ml + 9 ml = 10 ml. In this way, the principal weakened fixation is:
C₁= 1.33 M * 1 ml/10 ml
= 1.33 M * 1/10
= 0.133 M
The second dilution,
It is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M *1 ml/10 ml
= 0.133 M * 1/10
= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 5
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
Final concentration = 1.33 x 10⁻⁵ M
Find more information about Law of dilution here:
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