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what volume of a 3.95 m potassium chloride solution would be needed to …

Question

what volume of a 3.95 m potassium chloride solution would be needed to make 325 ml of a 2.76 m solution by dilution? ? ml of 3.95 m kcl

Explanation:

Step1: Recall dilution formula

The dilution formula is \( M_1V_1 = M_2V_2 \), where \( M_1 \) is the initial molarity, \( V_1 \) is the initial volume, \( M_2 \) is the final molarity, and \( V_2 \) is the final volume.

Step2: Identify known values

We know \( M_1 = 3.95 \, M \), \( M_2 = 2.76 \, M \), and \( V_2 = 325 \, mL \). We need to find \( V_1 \).

Step3: Rearrange formula to solve for \( V_1 \)

Rearranging \( M_1V_1 = M_2V_2 \) gives \( V_1=\frac{M_2V_2}{M_1} \).

Step4: Substitute values into formula

Substitute \( M_2 = 2.76 \, M \), \( V_2 = 325 \, mL \), and \( M_1 = 3.95 \, M \) into the formula: \( V_1=\frac{2.76\times325}{3.95} \).

Step5: Calculate the value

First, calculate \( 2.76\times325 = 897 \). Then, divide by \( 3.95 \): \( V_1=\frac{897}{3.95}\approx227.09 \, mL \).

Answer:

\( \approx 227 \) (or more precisely \( 227.09 \)) mL