Concentration has many units but the most commonly employed unit is Molarity. It is defined as the total number of moles of a solute that are dissolved in a liter of a solution. The method for calculating molarity is not so difficult rather it is a quite easy task and little practice can make you an expert in it.

The most important thing for calculating molarity can be derived simply by the unit of molarity i.e. moles of solute in addition to the liters of solution.

A very trouble-free example is given here which is solved with a very easy method and it covers almost all the steps that a molarity problem can suffer from:

Question: 10 grams of NaHCO3 are sufficient to make a saturated solution containing the volume of 500 milli-liters. Calculate molarity of this solution?

Solution:

If we closely observe this example, then it does not contain either of the parameters that are written in the formula of molarity i.e. moles of solute and liters of solution. But here weight of solute is given in grams and volume of solution is given in milli-liters.

The most important thing for calculating molarity can be derived simply by the unit of molarity i.e. moles of solute in addition to the liters of solution.

A very trouble-free example is given here which is solved with a very easy method and it covers almost all the steps that a molarity problem can suffer from:

Question: 10 grams of NaHCO3 are sufficient to make a saturated solution containing the volume of 500 milli-liters. Calculate molarity of this solution?

Solution:

If we closely observe this example, then it does not contain either of the parameters that are written in the formula of molarity i.e. moles of solute and liters of solution. But here weight of solute is given in grams and volume of solution is given in milli-liters.

First we need to find out the number of moles: molar mass is needed for the conversion of grams of solute into its number of moles. Molar mass can be obtained from the periodic table of elements.

(Molar Mass = Mm)

Mm of Na = 23.0 g

Mm of H = 1.0 g

Mm of C = 12.0 g

Mm of O = 16.0 g

Now from these individual elements, we can calculate the molar mass of the compound i.e. NaHCO3

Mm of NaHCO3 = 23.0 g + 1.0 g + 12.0 g + (16.0 g x 3)

Mm of NaHCO3 = 23.0 g + 1.0 g + 12.0 g + 48.0 g

Mm of NaHCO3 = 84.0 grams

From this, we can calculate moles as,

Moles = mass in grams / molecular mass

Moles = 10 / 84.0

Moles = 0.1190 moles

Now, the 2nd target is liters of the solution. It should be noted that it is the volume of whole solution rather than the solution of merely the solvent. As in this example, milli-liters of solution are given. So now these will be converted into the liters of the solution.

Conversion of 500 ml into liters:

Liters of solution = ml of the solution x (1 liter / 1000 milli-liters)

Liters of solution = 500 ml of the solution x (1 liter / 1000 milli-liters)

Liters of solution = 0.5 ml of the solution x (1 liter / 1000 milli-liters)

Now all the data for calculating molarity is completed. So now the next step will be calculation of molarity:

Molarity = number of moles of solute / liters

Molarity = 0. 1190 moles of NaHCO3 / 0.5 liters.

Molarity = 0.238 M

The required molarity of the given solution is calculated by the method as 0.238 M.

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