Name ___________________________________________ Date ______________ Period __________
Molarity Practice
Manipulating the Equation for Molarity:
Now that you’ve had adequate practice using this equation just to solve for molarity, we are
going to start manipulating this equation to solve for the other variables, given molarity.
Remember, plug in what you know, use “x” + units as the variable to solve for your unknown.
Molarity (M) =
moles of solute (mol)
Liters of solution (L)
1. How many moles of the following solutes would you need to prepare the indicated volume
and concentration of the solutions given?
a. 340 mL of a 1.82 M aluminum nitrate, Al(NO3)3 solution
• Convert to the necessary units that are used in the equation first (mL → L)
• We are going to have to solve for moles in this equation, because we are given
Molarity and Liters (after we converted). Use your algebraic skills to solve for “x”
moles.
!
1.82M =
x moles
; solve for x moles (show your work below)
______ L
b. 500 mL of a 5.2 M silver nitrate, AgNO3 solution
c. 375 mL of a 2.5 M ammonium nitrate, NH4NO3 solution
2. How many grams of the following solutes would you need to prepare the indicated
volume and concentration of the solutions given? (HINT: solve for moles first, then convert
to grams)
a. 325 mL of a 3.82 M magnesium nitrate, Mg(NO3)2 solution
• Convert to the necessary units that are used in the equation first (mL → L)
!
325 mL = ______________ L
• We are going to have to solve for moles in this equation first, because we are given
Molarity and Liters (after we converted). Use your algebraic skills to solve for “x”
moles, THEN convert moles to grams (you will need the molar mass of your
substance)
!
3.82M =
x moles
; solve for x moles (show your work below)
______ L
• Convert your moles you just solved for to GRAMS of your solute using the molar
mass
______ mole Mg(NO 3 )2 ×
______ g Mg(NO 3 )2
= _________ g Mg(NO 3 )2
1 mole Mg(NO 3 )2
b. 450 mL of a 2.2 M sodium nitrate, NaNO3 solution
c. 325 mL of a 3.45 M ammonium sulfide, (NH4)2S solution
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3. What would the final volume (in Liters) of each solution be so that the amount of solute
dissolved will produce the indicated concentration?
a. 2.86 g of copper (I) carbonate, Cu2CO3, to produce 0.640 M solution
• Convert to the necessary units that are used in the equation first (g → moles)
!
2.86 g Cu 2CO 3 ×
1 mole Cu 2CO 3
= _________ mole Cu 2CO 3
_______ g Cu 2CO 3
• We are going to have to solve for L in this equation because we are given Molarity
and moles (after we converted). Use your algebraic skills to solve for “x” Liters.
!
0.640M =
______ moles
; solve for x L (show your work below)
xL
b. 10.52 g of barium phosphate, Ba3(PO4)2 to produce 5.4 M solution
c. 55.25 g of iron (II) sulfate, FeSO4 to produce 2.75 M solution
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4. What will be the final concentration (M) of a solution prepared by dissolving the indicated
solute in enough water to produce the indicated volume of solution?
a. 15.4 g of sodium sulfate, Na2SO4, filled up to 340 mL
• Convert to the necessary units that are used in the equation first (g → moles)
!
15.4 g Na 2SO 4 ×
1 mole Na 2SO 4
= _________ mole Na 2SO 4
_______ g Na 2SO 4
!
!
340 mL = ______________ L
• We are going to have to solve for Molarity (M) in this equation because we are given
Liters and moles (after we converted to both). Use your algebraic skills to solve for
“x” Molarity (M).
!
xM=
______ moles
; solve for x M
_______ L
b. 25.43 g of sodium nitride, Na3N filled up to 155 mL
c. 65.45 g of vanadium (IV) phosphate, V3(PO4)4 filled up to 1025 mL
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