# Question a1580

Oct 11, 2016

$\text{349 mL}$

#### Explanation:

The molarity of the solution essentially tells the concentration of the solute in terms of moles per liter of solution.

In other words, a solution's molarity tells you how many moles of solute you get in exactly $\text{1 L}$ of solution. In your case, a $\text{0.150 M}$ copper(I) nitrate solution will contain $0.150$ moles of copper(I) nitrate for every liter of solution.

Now, use the molar mass of the compound to calculate how many moles you have in that sample

6.58 color(red)(cancel(color(black)("g"))) * "1 mole CuNO"_3/(125.55color(red)(cancel(color(black)("g")))) = "0.05241 moles CuNO"_3

Now all you have to do is use the molarity of the solution to figure out the volume needed to provide that many moles of solute

0.05241 color(red)(cancel(color(black)("moles CuNO"_3))) * "1 L solution"/(0.150 color(red)(cancel(color(black)("moles CuNO"_3)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.349 L")color(white)(a/a)|)))

The answer is rounded to three sig figs.

If you want, you can convert the volume to milliliters

0.349 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "349 mL"#