How many mL of a 0.250 M K₂SO₄ solution is required to prepare 500 mL of a 0.080 M K₂SO₄ solution?

Question

Quizplus · Accepted Answer

To find the volume of the original solution needed, use the dilution formula $C_1V_1 = C_2V_2$, 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 final solution. Plugging in the given values: $0.250 \, \text{M} \times V_1 = 0.080 \, \text{M} \times 500 \, \text{mL}$. Solving for $V_1$ gives $V_1 = \frac{0.080 \, \text{M} \times 500 \, \text{mL}}{0.250 \, \text{M}} = 160 \, \text{mL}$.

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