# If 3.38 g of CuNO_3 is dissolved in water to make a 0.450 M solution, what is the volume of the solution?

Feb 6, 2017

I will assume you mean $\text{cupric nitrate}$, $C u {\left(N {O}_{3}\right)}_{2}$.

#### Explanation:

$\text{Concentration"="Moles of solute"/"Volume of solution}$

And thus $\text{Volume of solution"="Moles of solute"/"Concentration}$

$= \frac{3.38 \cdot g}{187.56 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{0.450 \cdot m o l \cdot {L}^{-} 1} = 0.040 \cdot L$

Are you happy with this treatment? All I have done is manipulated the quotient, to give me an answer in terms of $\text{volume}$.

Note that if they really asked for $\text{cuprous nitrate}$, $C u N {O}_{3}$, they should have known better. I doubt that you could isolate $\text{cuprous nitrate}$.