Concentration Calculation Examples | Molarity

In this tutorial, we study about molarity and molality theories of solutions and solve questions to get better understand.



Definition of concentration

Concentration tells us, the amount of solute present in a given quantity of solvent or solution is expressed in terms of concentration. We can express concentration in different ways as below. units of those parameters are noted.

  1. Mass percentage (m/m%) or volume percentage (v/v%) - No units
  2. Molarity of solution - mol dm-3
  3. Molality of solution - mol kg-1
  4. Mole fraction - No units

Solvent, solute, solution definition

What is solvent?

The component is present in greater amount


What is the solute?

The component is present in lesser amount.


What is solution?

The mixture of solvent and solute


solvent solute solution

In this tutorial we try to study the molarity and molality of solutions.


Molarity of a solution

The number of moles of the solute dissolved per unit volume of the solution is the definition of molarity. Units can be mol dm-3 or mol m-3.


Molarity equation

molarity equation

Molarity units

Molarity can be expressed in different units.

  • mol dm-3 : mostly used
  • mol l-
  • kmol m-3


How to calculate concentration in mol dm-3 ?

In the units, there are mol and dm-3. They tells us about amount and volume respectively. Therefore we have to find the dissolved amount and volume of the solution to calculate concentration in mol dm-3.


Examples about molarity values

  • If 1 dm3 aqueous solution contains 0.45 mol of HCl, molarity of HCl is 0.45 mol dm-3. Otherwise, we can describe this is as, 0.45 mol of HCl is dissolved 1 dm3 of the total solution.
  • If 10 dm3 aqueous solution contains 0.98 mol of HNO3, molarity of HNO3 is 0.098 mol dm-3.

Sometimes we use '[ solute ]' to represent concentration of solute. As an example, to represent concentration of aqueous HCl, we can use [HCl(aq)]

Example: [HNO3(aq)] = 0.1 mol dm-3



Calculating molarity of solutions

Before calculating molarity of solutions, you should have an idea about following parametres.

  • Calculating molar mass when relative atomic masses are known
  • Relationship of molar mass (M), mass (m) and amount (n): n = m/M


Calculate molarity - Example 1



5.85 g of NaCl is dissolved in 500cm3 of distilled water. Calculate the molarity of NaCl mol dm-3. ( Na = 23 , Cl = 38.5 )

Answer

Dissolved amount (moles) and totoal volume of the solution are required to calculate the concentration according to the molarity equation. In this example, total solution volume is given as a data. Therefore we have to find the dissolved NaCl amount.



Find the amount of moles of dissolved NaCl

Mass of dissolved NaCl is given. Now we can find the NaCl amount using mass, molar mass and amount relationship. First molar mass of NaCl should be calculated.

Let's calculate the molar mass of NaCl

molar mass of NaCl = 23 *1 + 35.5*1

molar mass of NaCl = 58.5 g mol-1


Amount = mass / molar mass


Dissolved NaCl amount = 5.85 g / 58.5 g mol-1

Dissolved NaCl amount = 0.1 mol


Calculate the volume of total solution in dm3

1dm3 = 1000cm3

500cm3 = 500 * 10-3 dm3 = 0.5dm3

Now, we know the volume of solution in dm3 and amount of moles of NaCl. So we can find molarity of NaCl now.


Molarity Equation

molarity definition

Substitute NaCl moles and volume of the solution.

molarity example problem

Molarity of NaCl = 0.2 mol dm-3




Example 2

You are advised to prepare 0.47 mol dm-3 NaOH solution in the laboratory. Solid NaOH, distilled water and other lab facilities are supplied for you. If you have to prepare 25cm3 solution, calculate followings.

  1. Required amount of moles of NaOH
  2. Required NaOH mass

In this example, required molarity 0.47 mol dm-3 and volume are given. When we know the molarity and volume, we can easily calculate the required amount of moles of NaOH using relationship of concentration, moles, and volume of solution. After calculating NaOH moles, required NaOH mass is calculated.


Calculate moles of NaOH

Concentration = Moles / Volume

Moles of NaOH = Concentration * volume

Moles of NaOH = 0.47 mol dm-3 * 25 * 10-3 dm3

Moles of NaOH = 0.01175 mol


Calculate required mass of NaOH

Now we can use the relation of mass, molar mass and amount of moles to find required mass of NaOH.

Molar Mass of NaOH = 23 + 16 + 1

Molar Mass of NaOH = 40 g mol-1

Amount = Mass / Molar Mass

Required NaOH mass = NaOH amount * Molar Mass of NaOH

Required NaOH mass = 0.01175 mol * 25 * 40 g mol-1

Required NaOH mass = 0.47 g



Example 3

A mixture of Na2CO3 and K2CO3 is dissolved in distilled water upto 250 cm3. Dissolved mixture includes 10.6 g of Na2CO3 and 13.8 g of K2CO3.

  1. Calculate amount of moles each Na2CO3 and K2CO3 compounds.
  2. Express concentrations in mol dm-3 of each Na2CO3 and K2CO3.
  3. Calculate concentrations of CO32- in the solution.

K=39 , Na=23 , O=16 , C=12


1.

Masses of Na2CO3 and K2CO3 are given as data. So we can find amount of Na2CO3 and K2CO3. In this example, We use 'M' to express molar mass as a symbol.


Find molar masses of Na2CO3

MNa2CO3 = 23*2 + 12 + 16*3

MNa2CO3 = 106 g mol-1


Find amount of Na2CO3

Amount of Na2CO3 = 10.6 g / 106 g mol-1

Amount of Na2CO3 = 0.1 mol




Find molar masses of K2CO3

MK2CO3 = 39*2 + 12 + 16*3

MK2CO3 = 138 g mol-1


Find amount of K2CO3

Amount of K2CO3 = 13.8 g / 138 g mol-1

Amount of K2CO3 = 0.1 mol




2.

Now we know the amount of moles of each Na2CO3 and K2CO3. Therefore molarity can be found now.


Find molarity


Molarity = Dissolved Amount / Volume of Solution


[Na2CO3(aq)] = 0.1 mol / 0.25 dm3

[Na2CO3(aq)] = 0.4 mol dm-3



[K2CO3(aq)] = 0.1 mol / 0.25 dm3

[K2CO3(aq)] = 0.4 mol dm-3




3.

CO32- ions are given from both Na2CO3 and K2CO3 compounds. Both Na2CO3 and K2CO3 dissociate completely to ions in the water. Therefore total amount of CO32- ions in the water equals to the sum of CO32- given by Na2CO3 and CO32- given by K2CO3. After calculating total CO32-, molarity is calculated.

sodium carbonate in water

Total CO32- amount = CO32- from Na2CO3 + CO32- from K2CO3

Total CO32- amount = 0.1 + 0.1

Total CO32- amount = 0.2 mol


Therefore [CO32-] = 0.2 mol / 0.25 dm3

Therefore [CO32-] = 0.8 mol dm-3



Example 4

You have to prepare a NaOH solution and its pH value should be 13. Total volume of the solution is 500 ml. Calculate the required mass of NaOH to prepare this solution at 250C.

Calculate concentration of a pH known solution

Calculate pOH from pH by using the pH + pOH = 14 (at 250C)
Then you can calculate the concentration of NaOH solution by pOH = -log10[OH-(aq)] .
Then find the required moles of NaOH by the equation of C=n/v . Here C= concentration, n=required moles, v-volume of solution
Now weight is measured by multiplying n and molar mass.


Assumptions

The OH- received from water dissociation is negligible when it compare with OH- received from NaOH.


calculate weight concentration of solution

Calculate pOH from pH value

pH + pOH = 14
13 + pOH = 14
pOH = 1


Calculate concentration(mol dm-3)of solution

Now we know the pOH of NaOH solution. Use the relation of pOH and OH- concentration

pOH = -log10[OH-(aq)]
1 = -log10[OH-(aq)]
[OH-(aq)] = 0.1 mol dm-3


Find the required moles(mol) of NaOH

Concentration, moles amount, and volume relation is used.

C = n/v
0.1 = n/0.5      [500ml = 0.5 dm3]
n = 0.05mol


Find the required weight(g)

Multiply required moles times molecular weight

m = mass
n = moles
M = molecular weight
n = m/M

m = M*n
m = 0.05 * 40
m = 2g




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