Molarity is also called amount concentration, molar concentration or substance concentration. It is a measure of the concentration of a chemical species, in particular of a solute in a solution. In terms of the amount, it is a substance per unit volume of solution. In chemistry, the most commonly used unit for the term molarity is the number of moles per litre. A solution with a concentration of 1 mol/L is said to be 1 molar. It is commonly designated as 1 M.

In simple language, the molarity of a given solution is the total number of moles of solute per litre of solution. The molality of a solution depends on the changes in the physical properties of the system. These properties can be pressure and temperature unlike, mass. The volume of the system changes with the change in the physical conditions of the system. One molar is the molarity of a solution when one gram of solute is dissolved in a litre of solution. Most importantly, in a solution, the solvent and solute blend to form a solution. Hence, the total volume of the solution is taken.

## Unit of Molarity

According to the International System of Units (SI), the coherent unit for molarity or molar concentration is \(mol/m^3\). However, this is inconvenient for most laboratory purposes and most of the chemical literature traditionally uses mol/dm^{3}. This is the same as mol/L. This traditional unit is often represented by the letter M, optionally preceded by an SI prefix mega as needed to denote sub multiples. For instance,

\(mol/m^3= 10^-3 mol/dm^3 = 10^-3 mol/L = 10^-3 M = 1 mmol/L = 1 mM\)

The units millimolar and micromolar are referring to mM and \(\mu M (10^-3 mol/L and 10^-6 mol/L)\), respectively.

### Molarity Formula

The equation to calculate the 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

V is the volume of solution given in terms of litres

### Solved Example of Molarity

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

Solution: The molecular formula for sodium sulphate is \(Na_2SO_4\).

The molecular formula for water is \(H_2O\).

The molecular mass of sodium sulphate is

\(M = 23 \times 2 + 32 + 16 \times 4=142\)

The number of moles of sodium sulphate as per the given question is calculated by the formula,

\(N = mass in grams / molecular weight = \frac{15}{142} =0.106\)

Given that, 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 below, calculate the molarity of the given solution.

Molarity = number of moles of solute/volume of solution in litres

Substituting the values, we get,

Molarity = \(\frac{0.016}{0.125} = 0.85 mol/L\)

The molarity of the given solution is 0.85M or 0.85 mol/L.

## FAQs on Molarity

Question 1: How is molarity different from molality?

Answer: Molarity, which is denoted by M, is defined as the number of moles of solute present in one litre of solution. It is:

Molarity (M) = \(\frac{number of moles of solute (n)}{Volume of solution in liters (V)}\)

Whereas, Molality is denoted by m. Molality is defined as the number of moles of solute present per kilogram of the solvent.

The formula of molality is:

Molality (m) = \(\frac{number of moles of solute (n)}{Volume of solution in kg (V)}\)

Question 2: What is the definition of molarity?

Answer: Molarity (M) indicates the number of moles of solute in one litre of solution. Its unit is moles/Liter. Mathematically, it is:

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

It is the ratio of the moles of solute by the number of litres of its solution. With the molarity calculation, it is easy to identify the exact amount present of any given element or compound in a solution. In simple language molarity of any given solution is a method for knowing the specific elements or compounds that are present in any given solution.

Question 3: A solution is prepared by bubbling 1.56 grams of hydrochloric acid in water. The volume of the solution is 26.8 ml. Find the molarity of the solution.

Answer: The chemical formula of hydrochloric acid = HCl

The chemical formula for Water = \(H_2O\)

The molecular weight of HCl = \(35.5\times 1 + 1\times 1 = 36.5 moles/gram\)

The molecular weight of \(H_2O = 1\times 2 + 16\times 1 = 18 moles/gram\)

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

The number of moles of hydrochloric acid,

n = mass in grams / molecular weight

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

Now, volume of the solution = 26.8 mL

Expressing the volume in terms of litres,

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

Now calculate the molarity of the solution using the formula

Molarity = Number of moles of element / volume of solution in litres

\(Molarity = \frac{4.27\times 10^{-2}}{2.68 \times 10^{-2}} = 1.59 mol/L\)

The molarity of the solution is 1.59 M or 1.59 mol/L.

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