Interpretation:
Volume of 3.0 M KI stock solutionis to be calculated to prepare 0.3 L of 1.25 M KIsolution.
Concept introduction:
In dilution, the amount of solute does not change, the number of moles is the same before and after dilution.
If subscript "1" represents initial and "2" represents the final values of the quantities involved, we have:
Here, molarity is used as a unit of concentration.
The equation for dilution is M1V1 = M2V2
Stock solution = Diluted solution
Where
M1 = molarity of the stock solution
V1 = volume of stock solution
M2 = molarity of the diluted solution
V2 = volume of diluted solution
Answer to Problem 24PP
125 mL of 3.0 M KI stock solutionis required to prepare 0.3 L of 1.25 M KIsolution.
Given information:
The molarity of the stock solution M1 = 3.0 M The molarity of the diluted solution M2 = 1.25 M The volume of diluted solution V2 = 0.3 L
= 300 mL
Explanation:
The equation for dilution is as follows:
Here,
M1 = molarity of the stock solution,
V1 = volume of stock solution,
M2 = molarity of the diluted solution,
V2 = volume of diluted solution.
Volume of stock KI solution =
Thus, 125 mL of 3.0 M KI stock solutionis required to prepare 0.3 L of 1.25 M KIsolution.
Explanation of Solution
Given information:
The molarity of the stock solution M1 = 3.0 M The molarity of the diluted solution M2 = 1.25 M The volume of diluted solution V2 = 0.3 L
= 300 mL
The equation for dilution is as follows:
Here,
M1 = molarity of the stock solution,
V1 = volume of stock solution,
M2 = molarity of the diluted solution,
V2 = volume of diluted solution.
Volume of stock KI solution =
Thus, 125 mL of 3.0 M KI stock solutionis required to prepare 0.3 L of 1.25 M KIsolution.
Chapter 14 Solutions
Chemistry: Matter and Change
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