An aqueous solution is prepared by dissolving 15.6 g of Cu(NO3)2 ⋅ 6 H2O in water and diluting to 345 mL of solution. What is the molarity of the NO3- ion in this solution?

A. 0.417 M
B. 0.121 M
C. 0.306 M
D. 0563 M
E. 0.834 M

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Mass of solute [tex](Cu(NO_3)_2.6H_2O)[/tex] = 15.6 g

Molar mass of [tex](Cu(NO_3)_2.6H_2O)[/tex] = 295.6 g/mol

Volume of solution = 345 mL

Putting values in above equation, we get:

[tex]\text{Molarity of }Cu(NO_3)_2.6H_2O=\frac{15.6g\times 1000}{295.6g/mol\times 345mL}\\\\\text{Molarity of }Cu(NO_3)_.6H_2O=0.153M[/tex]

As, 1 mole of [tex](Cu(NO_3)_2.6H_2O)[/tex] produces 1 mole of copper (II) ions and 2 moles of nitrate ions.

So, molarity of [tex]NO_3^-[/tex] ions = (2 × 0.153) = 0.306 M

Hence, the molarity of [tex]NO_3^-[/tex] ions in the solution is 0.306 M

An aqueous solution is prepared by dissolving 15.6 g of Cu(NO3)2 ⋅ 6 H2O in water and diluting to 345 mL of solution. What is the molarity of the NO3- ion in this solution? A. 0.417 M B. 0.121 M C. 0.306 M D. 0563 M E. 0.834 M

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An aqueous solution is prepared by dissolving 15.6 g of Cu(NO3)2 ⋅ 6 H2O in water and diluting to 345 mL of solution. What is the molarity of the NO3- ion in this solution?

A. 0.417 M
B. 0.121 M
C. 0.306 M
D. 0563 M
E. 0.834 M