Question

What mass of `Mg(NO_3)_2` is present in 145 mL of a .150 M solution of `Mg(NO_3)_2`?

Answer 1

`"3.23 g"`

Explanation:

The idea here is that you need to use the molarity and volume of the solution to figure out how many moles of magnesium nitrate, `"Mg"("NO"_3)_2`, you have in that sample.

To do that, use the molarity of the solution as a conversion factor, but keep in mind that you must also convert the volume of the solution from milliliters to liters

`145 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.150 moles Mg"("NO"_ 3)_ 2)/(1color(red)(cancel(color(black)("L solution")))`

`= "0.02175 moles Mg"("NO"_3)_2`

Now all you have to do is to convert the number of moles of magnesium nitrate to grams by using the molar mass of the compound

`0.02175 color(red)(cancel(color(black)("moles Mg"("NO"_3)_2))) * "148.3 g"/(1color(red)(cancel(color(black)("mole Mg"("NO"_3)_2)))) = color(darkgreen)(ul(color(black)("3.23 g")))`

The answer is rounded to three sig figs.