What mass of #Mg(NO_3)_2# is present in 145 mL of a .150 M solution of #Mg(NO_3)_2#?
1 Answer
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,
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.
