What is the molarity of a solution when 75.0 mL of a 4.00 M KCl solution is diluted to a volume of 500. mL?

Question

Accepted Answer

STEP 1 Prepare a table of the concentrations and volumes of the solutions.

ANALYZE THE PROBLEM

Given

Need

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[latex]C_{1} = 4.00 M V_{1} = 75.0 mL V_{2} = 500. mL[/latex]

[latex] \boxed{C_{2}}[/latex]

[latex] C_{1}V_{1} = C_{2}V_{2}[/latex]

Predict: [latex]V_{2}[/latex] increases,
[latex]C_{2}[/latex] decreases

STEP 2 Rearrange the dilution expression to solve for the unknown quantity.

[latex] C_{1} V_{1}= \boxed{C_{2}}V_{2} [/latex]

[latex] \frac{C_{1}V_{1}}{V_{2}} = \frac{\boxed{C_{2}}V_{2}}{V_{2}} [/latex] Divide both sides by [latex] V_{2} [/latex]

[latex] \boxed{C_{2}}= C_{1} imes \frac{V_{1}}{V_{2}}[/latex]

STEP 3 Substitute the known quantities into the dilution expression and calculate.

[latex] \boxed{C_{2}} =\underset{Three SFs}{4.00 M} imes \overset{Three SFs}{\underset{decreases concentration }{\underset{Volume factor }{\underset{Three SFs}{\frac{75.0 \cancel{mL}}{500. \cancel{mL}}} } }} = \underset{Three SFs}{0.600 M (diluted KCl solution) }[/latex]

When the initial molarity [latex](C_{1})[/latex] is multiplied by a ratio of the volumes (volume factor) that is less than 1, the molarity of the diluted solution [latex](C_{2})[/latex] decreases as predicted in Step 1.

Question 9.14: What is the molarity of a solution when 75.0 mL of a 4.00 M ...

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