**Definition:**Molarity of a given solution is defined as the total number of moles of solute per litre of solution. The molality of a solution is dependent on the changes in physical properties of the system such as pressure and temperature as unlike mass, the volume of the system changes with the change in physical conditions of the system. Molarity is represented by M, which is termed as molar. One molar is the molarity of a solution where one gram of solute is dissolved in litre of solution. As we know, in a solution, the solvent and solute blend to form a solution, hence, the total volume of the solution is taken.

**Molarity Formula:**

The equation for calculating molarity is the ratio of the moles of solute whose molarity is to be calculated and the volume of solvent used to dissolve the given solute.

\(M = \frac{n}{V}\)

Here, M is the molality of the solution that is to be calculated, n is the number of moles of the solute and V is the volume of solution given in terms of litres.

**Example 1:**

A solution is prepared by bubbling 1.56 grams of hydrochloric acid in water. Here, the volume of the solution is 26.8 mL. Calculate the molarity of the solution.

**Solution: **

The chemical formula of hydrochloric acid = HCl

The chemical formula for Water = H2O

The molecular weight of HCl = 35.5 Ã—1 + 1Ã—1 = 36.5 moles/gram

The molecular weight of H2O = 1 Ã—2 + 16 Ã—1 = 18 moles/gram

Given, mass of hydrochloric acid in the solution = 1.56 g

The number of moles of hydrochloric acid =

\(n_{T}=\frac{mass\;in\;grams}{molecular\;weight}\)

\(n_{T}=\frac{1.56}{36.5}=4.27\times 10^{-2}mole\)

Now, given volume of the solution = 26.8 mL

Expressing the volume in terms of litres,

\(volume = \frac{26.8}{1000}=2.68\times 10^{-2}Litre\)

Now, we calculate the molarity of the solution using the formula given above.

\(Molarity = \frac{Number\;of\;moles\;of\;element}{volume\;of\;solution\;in\;litres}\)

\(Molality = \frac{4.27\times 10^{-2}}{2.68\times 10^{-2}}=1.59M\)

The molarity of the solution is 1.59 M.

**Example 2: **

A solution prepared using 15 g of sodium sulphate. The volume of the solution is 125 ml. Calculate the molarity of the given solution of sodium sulphate.

**Solution: **

The molecular formula for sodium sulphate is Na2SO4.

The molecular formula for water is H2O.

The molecular mass of sodium sulphate is calculated as given below,

M=23Ã—2+32+16Ã—4=142

The number of moles of sodium sulphate in the given question is calculated as,

\(n=\frac{mass\;in\;grams}{molecular\;weight}=\frac{15}{142}=0.106\)

The volume of the solution is 125 ml.

Expressing the above values in terms of litres,

\(Volume = \frac{125}{1000}=0.125\)

Now, using the formula given above, we calculate the molarity of the given solution.

\(molarity = \frac{number\;of\;moles\;of\;solute}{volume\;of\;solution}\)

Substituting the values, we get,

\(molarity=\frac{0.106}{0.125}=0.85M\)

The molarity of the given solution is 0.85M.