Question

If 3.58 g CuNO3is dissolved in water to make a 0.670 M solution, what is the volume of the solution?

Answer 1

Using equation for molar concentration $c=\frac{n}{V}$

, where c - molar concentration, n - amount of compound in moles, V - volume of solution

we can determine volume of solution:

$V(CuNO_3)=\frac{n(CuNO_3)}{c(CuNO_3)}$

The amount n(CuNO₃) can be found from the equation

$n(CuNO_3)=\frac{m(CuNO_3)}{M(CuNO_3)}$

,where m - mass of CuNO₃ and M(CuNO₃) is molar mass of CuNO_3.

Thus $V(CuNO_3)=\frac{n(CuNO_3)}{c(CuNO_3)}=\frac{m(CuNO_3)}{M(CuNO_3) \cdot c(CuNO_3)}=\frac{3.58g}{126\frac{g}{mol} \cdot 0.670 \frac{mol}{L}}=0.042L=42mL$

By the way, copper (I) nitrate CuNO₃ doesn't exist in pure form without ligands, thus the problem in the given wording doesn't make any practical sense.