A laboratory procedure calls for making 453.3 ml of a 2.62M salt solution. If the molar mass of the salt is 218 g/mol what mass is required?

Jul 8, 2016

$\text{259 g}$

Explanation:

Your strategy here will be to

• use the molarity and volume of the solution to determine how many moles of solute it must contain
• use the molar mass of the salt to calculate how many grams would contain that many moles

A solution's molarity tells you how many moles of solute you have in one liter of solution.

In your case, the solution must have a molarity of ${\text{2.62 mol L}}^{- 1}$, which means that every liter of solution will contain $2.62$ moles of solute.

Now, your sample is said to have a volume of $\text{453.3 mL}$. Use its molarity to figure out how many moles of solute it contains -- do not forget to convert the volume to liters!

453.3 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace("2.62 moles solute"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("known molarity")) = "1.188 moles solute"

As you know, a compound's molar mass tells you the mass of one mole of said compound. In your case, the salt has a molar mass of ${\text{218 g mol}}^{- 1}$, which means that one mole has a mass of $\text{218 g}$.

This means that the mass of $1.188$ moles will be

1.188 color(red)(cancel(color(black)("moles"))) * "218 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("259 g")color(white)(a/a)|)))

The answer is rounded to three sig figs.