## Molarity Calculation

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 CaCl2Insert 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)

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 :

We need 149.604 mL 1-propanol (in CHEMIX School propanol-1) (Procedure - dilution)

b)
******** Dilution by volume ********