Question

Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL. M solution

Answer 1

Answer:

$M_2=3.34x10^{-3}M$

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

$M_1V_1=M_2V_2$

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

$M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M$

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