## 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 = **n**_{solute}/ **V**_{solution}

**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 (CaCl_{2}).

Formula weight CaCl_{2} = 110.983g/mol

Solution**:**

**M** = **n**_{solute}/**V**_{solution}

**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 CaCl_{2} . *Insert mass
: ***45**

3) Insert volume **V**_{solu. } (mL) = **2000**

**Solution:** *0.202733M*

**Molarity Calculation Example 2**

A solution of H_{2}SO_{4} with a molal concentration
of 5.25**m** (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
H_{2}SO_{4} contain 0.514917Kg of H_{2}SO_{4}
and 1Kg of H_{2}O

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** = **n**_{solute}/**V**_{solution}
= 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 3**m** (molal) solution of ammonium
sulfate (NH_{4})_{2}SO_{4} 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=n_{solute}/V_{solution}
)

By knowing the density of the solution, the volume of
1.396423Kg can be calculated :

1.396423Kg /( 1.16308 Kg/L) = 1.20063L^{}

M = **n**_{solute}/V_{solution} =
3mol/1.2dm^{3} =** 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
(C_{3}H_{7}OH) (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(C_{3}H_{7}OH) = (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 -> V_{solu.}(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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