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What is the molar concentration of a solution prepared by dissolving 8.63 g of Ba(NO3)2 (molar mass = 261.35 g/mol) to a total volume of 250.0 mL?

What is the molar concentration of a solution prepared by diluting 10.0 mL of the solution in Question #1 to a total volume of 50.00 mL?

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The concentration of the solution is 0.132 M, the concentration of the diluted solution is  0.026 M.

Given that;

number of moles = concentration × Volume

Concentration = number of moles/volume

But number of moles = mass/molar mass

mass = 8.63 g

Molar mass =  261.35 g/mol

Hence, number of moles = 8.63 g/261.35 g/mol = 0.033 moles

volume =  250.0 mL or 0.25 L

concentration = 0.033 moles/0.25 L = 0.132 M

For the second part of the question;

Using the dilution formula, C1V1 =C2V2

C1 = initial concentration = 0.132 M

V1 = initial volume = 10.0 mL

C2 = Final concentration = ?

V2 = Final volume = 50.00 mL

C2 =  0.132 M × 10.0 mL/ 50.00 mL

C2 = 0.026 M

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