Question

Im hvaing a hard time getting the right answer

Answer 1

Answer:

$V=23.9mL$

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

$n=2.49g*\frac{1mol}{125.55 g}=0.0198mol$

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

$M=\frac{n}{V}\\\\V=\frac{n}{M}$

By plugging in the moles and molarity, we obtain:

$V=\frac{0.0198mol}{0.830mol/L}=0.0239L$

Which in mL is:

$V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL$

Best regards!