QuestionJuly 7, 2025

What volume of a 1.25 M potassium fluoride (KF) solution would be needed to make 455 mL of a 0838 M solution by dilution? () mL of 1.25 M KF

What volume of a 1.25 M potassium fluoride (KF) solution would be needed to make 455 mL of a 0838 M solution by dilution? () mL of 1.25 M KF
What volume of a 1.25 M potassium
fluoride (KF) solution would be needed
to make 455 mL of a 0838 M solution
by dilution?
() mL of 1.25 M KF

Solution
4.7(211 votes)

Answer

305 mL of 1.25 M KF Explanation 1. Use the dilution formula The formula for dilution is C_1V_1 = C_2V_2, where C_1 and V_1 are the concentration and volume of the initial solution, and C_2 and V_2 are the concentration and volume of the diluted solution. 2. Substitute known values Substitute C_1 = 1.25 \, \text{M}, C_2 = 0.838 \, \text{M}, and V_2 = 455 \, \text{mL} into the formula: 1.25 \times V_1 = 0.838 \times 455. 3. Solve for V_1 Rearrange to find V_1: V_1 = \frac{0.838 \times 455}{1.25}.

Explanation

1. Use the dilution formula<br /> The formula for dilution is $C_1V_1 = C_2V_2$, where $C_1$ and $V_1$ are the concentration and volume of the initial solution, and $C_2$ and $V_2$ are the concentration and volume of the diluted solution.<br />2. Substitute known values<br /> Substitute $C_1 = 1.25 \, \text{M}$, $C_2 = 0.838 \, \text{M}$, and $V_2 = 455 \, \text{mL}$ into the formula: $1.25 \times V_1 = 0.838 \times 455$.<br />3. Solve for $V_1$<br /> Rearrange to find $V_1$: $V_1 = \frac{0.838 \times 455}{1.25}$.
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