Question

A stock solution of Cu2+(aq) was prepared by placing 0.8875 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and diluting to the mark with water. What is the concentration (in M) of Cu2+(aq)in the stock solution? Molar mass of Cu = 63.55 g/mol.

Answer 1

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) was prepared by placing 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask and diluting to the mark with water. What is the concentration (in M) of Cu²⁺(aq)in the stock solution?

We can establish the following relations:

• The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
• 1 mole of Cu(NO₃)₂∙2.5H₂O contains 1 mole of Cu²⁺.

The moles of Cu²⁺ in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

$0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}$

The molarity of Cu²⁺ is:

$\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M$

Answer 2

The concentration of the $Cu^{2+}$ (aq) in the stock solution would be 0.03816 M

Molarity

Mole of 0.8875 g Cu(NO3)2∙2.5 H2O = 0.8875/232.59 = 0.003816 moles

Concentration of the solution = mole/volume (L) = 0.003816/0.1 = 0.03816 M

Since 1 mole Cu(NO3)2∙2.5 H2O contains 1 mole of $Cu^{2+}$ (aq), the concentration of $Cu^{2+}$ (aq) would also be 0.03816 M

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