# How many grams of "CuSO"_4 would be dissolved in 5.1 L of 0.5 M "CuSO"_4 solution?

## It is a molarity question.

Nov 21, 2016

$\text{410 g}$

#### Explanation:

Molarity is simply a measure of how many moles of solute you have in one liter of a given solution. In order for a solution to have a molarity of $\text{1 M}$, it must contain $1$ mole of solute for every $\text{1 L}$ of solution.

In your case, you have a $\text{0.5 M}$ solution of copper(II) sulfate, ${\text{CuSO}}_{4}$. This tells you that $\text{1 L}$ of this solution contains $0.5$ moles of copper(II) sulfate.

Since you have more than $\text{1 L}$ of this solution, you are going to get more than $0.5$ moles of solute. More specifically, you will get

5.1 color(red)(cancel(color(black)("L solution"))) * "0.5 moles CuSO"_4/(1color(red)(cancel(color(black)("L solution")))) = "2.55 moles CuSO"_4

Finally, use the molar mass of copper(II) sulfate to convert the number of moles to grams

$2.55 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CuSO"_4))) * "159.61 g"/(1color(red)(cancel(color(black)("mole CuSO"_4)))) = color(darkgreen)(ul(color(black)("410 g}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the molarity of the solution.