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A 35.00 mL sample of Mg(NO3)2 was diluted with water to 62.50 mL. A 20.00 mL sample of the dilute solution contained 0.300 M NO3−(aq). What was the concentration of Mg(NO3)2 in the original, undiluted solution? Hint:

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jufsanabriasa

Answer:

0,268M

Explanation:

A dilution is a process in which the concentration of a solution is reduced by adding more solvent.

In the problem, the sample of Mg(NO₃)₂ was diluted from 35,00mL to 62,50mL -dilution factor: 62,50mL/35,00mL- . The NO₃⁻ concentration of this diluted solution was 0,300M, as Mg(NO₃)₂ has two NO₃⁻, the concentration of Mg(NO₃)₂ in the diluted solution is:

0,300M NO₃⁻×[tex]\frac{1Mg(NO_{3})_2}{2NO_{3}^-}[/tex] = 0,150M Mg(NO₃)₂

Using dilution factor, the original concentration of Mg(NO₃)₂ is:

0,150M Mg(NO₃)₂×[tex]\frac{62,50mL}{35,00mL}[/tex] = 0,268M

I hope it helps!

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