Question

Calculate the concentration of Mg(NO3)2 in a solution prepared by diluting 81.00 mL of 0.217 M Mg(NO3)2 to a volume of 500.0 mL in a volumetric flask?

Answer 1

Answer on Question #41849, Chemistry, Other

Task:

Calculate the concentration of $\mathrm{Mg(NO_3)_2}$ in a solution prepared by diluting $81.00~\mathrm{mL}$ of 0.217 M $\mathrm{Mg(NO_3)_2}$ to a volume of $500.0~\mathrm{mL}$ in a volumetric flask?

Answer:

$$C _ {M} = \frac {v}{V} \quad v = \frac {m}{M}$$

where $C_M$ - molar concentration of a solution;

v - amount of moles of a certain substance;

m - mass of a substance, grams;

M - molar mass of a substance, g/mol.

The amount of moles in an initial solution of $\mathrm{Mg(NO_3)_2}$ is:

$$v = C _ {M} \cdot V$$ $$v \left(\mathrm{Mg} \left(\mathrm{NO}_{3}\right)_{2}\right) = 0.217 \cdot \frac{81.00}{1000} = 0.018 \text{ moles}$$

That is why the concentration of the solution after diluting in a volumetric flask will be:

$$C _ {M} \left(M g \left(N O _ {3}\right) _ {2}\right) = \frac{0.018}{0.500} = 0.036 M$$

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