# What is the Molarity of a Solution | Unit, Adavantages, eg

## Definition of Molarity

In chemistry, molarity is a concentration unit, defined to be the number of moles of solute divided by the number of litres of solution.

Molarity = no. of moles of solute /volume of solution in litres

The component of a solution that is present in the largest amount is known as the solvent. Any chemical species mixed in the solvent is called a solute, and solutes can be gases, liquids, or solids.

For example, Earth’s atmosphere is a mixture of 78% nitrogen gas, 21% oxygen gas, and 1% argon, carbon dioxide, and other gases.

We can think of the atmosphere as a solution where nitrogen gas is the solvent, and the solutes are oxygen, argon and carbon dioxide.

Chemists primarily need the concentration of solutions to be expressed in accounting for the number of particles that react according to a particular chemical equation.

Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions. A concentration unit based on moles is preferable.

• Molarity is the number of moles of solute dissolved per litre of the solution and one of the most common units used to measure the concentration of a solution. In terms of the amount of substance per unit volume of solution. In chemistry, the most commonly used unit for molarity is the number of moles per litre, having the unit symbol mol/L or mol⋅dm-3 in the SI unit.
• Molarity can be used to calculate the volume of solvent or the amount of solute.
• Molecular concentration is the number of molecules of a particular component per unit volume. Since the number of molecules in a litre or even a cubic centimetre is enormous, it has become common practice to use molar, rather than molecular, quantities.
• A mole is the gram-molecular weight of a substance and, therefore, Avogadro’s number ( Avogadro’s Law) of molecules (6.02×1023). Thus, the number of moles in a sample is the weight of the sample divided by the molecular weight of the substance; it is also the number of molecules in the sample divided by Avogadro’s number. Instead of using molecular concentration, it is more convenient to use molar concentration; For example, that the concentration is 12.04 × 1023 molecules per litre, it is simpler to say that it is two moles per litre. Concentration in moles per litre (i.e., molarity) is usually designated by the letter M. Molarity depends on the volume of the whole solution.

## Units of Molarity

Molarity is expressed in units of moles per litre (mol/L). It’s such a common unit, and it has its symbol, a capital letter M. A solution with the concentration of 5 mol/L would be called a 5M solution or said to have a concentration value of 5 molars.

## Examples of Molarity

• There are 6 moles of HCl in one litre of 6 molar HCl or 6 M HCl.
• There are 0.05 moles of NaCl in 500 ml of a 0.1 M NaCl solution. (The calculation of moles of ions depends on their solubility.)
• There are 0.1 moles of Na+ ions in one litre of a 0.1 M NaCl solution (aqueous).
• The molecular mass of Sulphuric acid is 98. If 98 grams of sulphuric acid is present in 1 litre of solution, then it is a 1 molar solution of Sulphuric acid. And therefore, molarity will be 1.

There are two big advantages of using molarity to express concentration.

• The first advantage is that it’s easy and convenient to use because the solute may be measured in grams, converted into moles, and mixed with a volume.
• The second advantage is that the sum of the molar concentrations is the total molar concentration. This permits calculations of density and ionic strength.

The big disadvantage of molarity is that it changes according to temperature.

This is because the volume of a liquid is affected by temperature. This is not a problem if measurements are all performed at a single temperature (e.g., room temperature).

However, it’s good practice to report the temperature when citing a molarity value. When making a solution, keep in mind, molarity will change slightly if you use a hot or cold solvent, yet store the final solution at a different temperature.

## Solved Examples of Molarity

Here are some solved examples of molarity;

### Example-1

Express the concentration of a solution of 1.2 grams of KCl in 250 ml of water.

Solution: To solve the problem, you need to convert the values into units of molarity, which are moles and litres. Start by converting grams of potassium chloride (KCl) into moles. To do this, look up the atomic masses of the elements on the periodic table. The atomic mass is the mass in grams of 1 mole of atoms.

mass of K = 39,10 g/mol

mass of Cl = 35.45 g/mol

So, the mass of one mole of KCl is:

mass of KCl = mass of K + mass of Cl

mass of KCl = 39.10 g + 35.45 g

Therefore, the mass of KCl = 74.55 g/mol

You have 1.2 grams of KCl, so you need to find how many moles that is:

moles KCl = (1.2 g KCl)(1 mol/74.55g)

moles KCl = 0.0161 mol

Now, you know how many moles of solute are present. Next, you need to convert the volume of solvent (water) from ml to L.

Remember. There are 1000 millilitres in 1 litre:

litres of water = (250 ml)(1 L/1000 ml)

litres of water = 0.25 L.

Finally, you’re ready to determine molarity. Express the concentration of KCl in water in terms of moles solute (KCl) per litres of solute (water):

molarity of solution = mol KC/L

water molarity = 0.0161 mol KCl/0.25 L water

molarity of the solution = 0.0644 M

Since you were given mass and volume using two significant figures, you should report molarity in 2 significant figures also:

molarity of KCl solution = 0.064 M

### Example-2

Determine the mass of NaOH dissolute for making a solution of 500 cm3, 12M molar.

Solution: As per the formula of molarity,

M = w/M × 1000/V ml

The molar mass of NaOH is = 23 + 16 + 1 = 40

Here,

M = 1/2 and V = 500 ml

Therefore,

1/2 = w/40 × 1000/500

Therefore,

W = 20/2 = 10g

### Example-3

Determine the molarity of NaOH solution which is prepared by dissolving its 4g amount in water to form a solution of 250 ml.

Solution: According to the formula of molarity

M = w/M × 1000/V ml = 4g/40g × 1000/250

= 0.1mol/0.250L

= 0.4 mol/L or 0.4M.