Question

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

Answer 1

Answer: The molarity of $NO_3^-$ ions in the solution is 0.306 M

Explanation:

To calculate the molarity of solution, we use the equation:

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

We are given:

Mass of solute $(Cu(NO_3)_2.6H_2O)$ = 15.6 g

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

Volume of solution = 345 mL

Putting values in above equation, we get:

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

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

So, molarity of $NO_3^-$ ions = (2 × 0.153) = 0.306 M

Hence, the molarity of $NO_3^-$ ions in the solution is 0.306 M