Answer:
[tex]V=23.9mL[/tex]
Explanation:
Hello!
In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:
[tex]n=2.49g*\frac{1mol}{125.55 g}=0.0198mol[/tex]
Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:
[tex]M=\frac{n}{V}\\\\V=\frac{n}{M}[/tex]
By plugging in the moles and molarity, we obtain:
[tex]V=\frac{0.0198mol}{0.830mol/L}=0.0239L[/tex]
Which in mL is:
[tex]V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL[/tex]
Best regards!