Molarity Calculation
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Molarity calculation examples
Molarity is the most common solution concentration in calculations
dealing with volumetric stoichiometry. The units of molarity, mol/L,
are represented by a scripted capital “M”. Molarity is
affected by temperature because it is based on the volume of the
solution, and the volume of a substance will be affected by changes
in temperature.
Molarity (M) = moles solute / liters of
solution = nsolute/ Vsolution
Solvent : the liquid in which a solute
dissolves
Solute : the substance that
dissolves in a liquid to form a solution
Solution : is the mixture formed when a solute has
dissolved in a solvent
Molarity Calculation Example 1
What is the molarity (M) of a 2L solution containing 45g of
dissolved calcium chloride (CaCl2).
Formula weight CaCl2 = 110.983g/mol
Solution:
M = nsolute/Vsolution
n = (45g)/(110.983g/mol) = 0.405468 mol
M = 0.405468mol / 2L = 0.202734M
CHEMIX School Solution:
1) Select substance: calcium chloride
2) We know the mass (g) of CaCl2 . Insert mass
: 45
3) Insert volume Vsolu. (mL) = 2000
Solution: 0.202733M
Molarity Calculation Example 2
A solution of H2SO4 with a molal concentration
of 5.25m (molal) has a density of 1.251 g/ml. what is the
molar concentration of this solution? Formula weight sulfuric acid
= 98.0795 g/mol
Solution:
Weight 5.25mol sulfuric acid = 98.0795g/mol *
5.25mol = 514.917 g
Assuming use of 1 Kg of water, a 5.25 molal solution of
H2SO4 contain 0.514917Kg of H2SO4
and 1Kg of H2O
Weight of solution = 0. 514917Kg H2SO4 + 1kg H2O = 1.514917Kg
Volume of 1.51491Kg solution = 1.514917Kg /
(1.251Kg/L) = 1.211L
Molarity of solution M = nsolute/Vsolution
= 5.25mol/1.211L = 4.334M
CHEMIX School Solution:
1) Select substance: sulfuric acid
2) Insert molal concentration (m): 5.25
Solution: 4.33682M
Molarity Calculation Example 3
The density of a 3m (molal) solution of ammonium
sulfate (NH4)2SO4 is 1.16308 g/mL.
Determine the molarity (M) of the
solution knowing that FW ammonium sulfate =
132.141g/mol.
Solution:
Assuming we use 1 Kg of water we kan make a 3molal solution by
adding 3*FW = 396.423g to 1 kg of water.
The total weight of the solution: 1Kg+0.396423Kg = 1.396423Kg
Molarity is related to volume ( M=nsolute/Vsolution
)
By knowing the density of the solution, the volume of
1.396423Kg can be calculated :
1.396423Kg /( 1.16308 Kg/L) = 1.20063L
M = nsolute/Vsolution =
3mol/1.2dm3 = 2.4987 M
CHEMIX School Solution:
1) Select substance: ammonium sulfate
2) Insert molal concentration (m): 3
Solution: 2.4987M
Molarity Calculation Example 4
a) How many mL of 1-propanol (100%) do we need to prepare 1000 mL of
a 2 M 1-propanol
Beregning av molaritet
(C3H7OH) (aq)solution.
Density 100% propanol-1 = 0.8034g/mL , FW=60.0959
b) How do we (in practice) make the 2M 1-propanol solution.
Solution:
a)
Calculate the volume of 2 moles of conc.(100%) 1-propanol
Beregning av molaritet
V(C3H7OH) = (n*FW) /d
= (2mol *60.0959 g/mol)/(0.8034g/mL) =
149.604mL
b)
Put about 800mL of water into a flask and add 149.604 mL
propanol-1.Add water until 1000 mL.
Beregning av molarit
CHEMIX School Solution:
a)
Select substance: propanol-1
In Dilution -> Stock concentration text field select w/w%
and insert: 100
Insert volume -> Amount -> Vsolu.(mL):
1000
Insert Amount -> M : 2
We need 149.604 mL 1-propanol (in CHEMIX School propanol-1)
(Procedure - dilution)
b)
******** Dilution by volume ********
Put about 600 mL of water into a flask and add 149.604 mL propanol-1
stock conc. Add water until 1000 mL
******** Dilution by mass
********
Put about 600 g of water into a flask and add 120.192 g of
propanol-1 stock conc. Add water until 980.083 g
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