You must be familiar with the sugar solution or solution of table salt in water. What are the difference and similarity in these two solutions? Yes, both solutions have one common component that is water. Do you know what solution is? The homogenous mixture of two components; solute and solvent are called as a solution. Here solute is present in less amount compared to solvent. Hence, in both of these solutions, water is solvent whereas sugar and table salt are solute. It is not necessary that only liquids can be solvent and solid can be solute. There are several examples of solutions in which solutes and solvents are found in different states like gas or solids.

For example; smoke is a solution of smoke particles in the air. Hence smoke particles are solute and air is solvent here. Other examples of solutions are foam, gem stone, butter, whipped cream etc. The concentration of solution can be defined as the amount of solute found in a certain volume of solution. There are several ways to express the concentration of solutions such as normality, molarity, molality, part per million, formality etc. In this article, we will discuss the molarity of the solution. Let's define molarity and learn how to determine the molarity of a solution which contains a certain amount of solute.

**Let us consider some problems related to molarity.**

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In current practice, concentration is mostly expressed as Molarity. Molarity can be defined as the **"number of moles of solute per 1 dm**^{3} or 1 liter o a solution". It is denoted by "M".

If n is the number of moles of solute and V is the volume in dm^{3} or liter, Molarity can be expressed as

Molarity is expressed, as it is defined.

- The substance dissolved, which is called a
**solute**. - The liquid in which it is dissolved called a
**solvent**.

There are several methods in which concentration of a solution can be expressed. Some of them are **Molarity, Molality and Normality**. Molarity is a simple and effective method of expressing the concentration of a solution.

Water is a solvent itself. Since we
usually calculate the amount of solute in a solvent, and refer to it as
concentration term, we cannot find the molarity of solvent alone.

**Mass of Solute from Molarity**

Mass of the solute can be found out from the molarity. The approach is reverse.In various chemical processes in industrial units, the mass of the solute is required to be found out from the molarity of supplied solution. It is also useful to calculate the mass of salts expected from sea water or other such procedures.

**For example,** Suppose the molarity of a solution of CaCl_{2} is o.1 M. Find the mass of CaCl_{2} in 100ml of the solution.

First find the no.of moles from the molarity and volume.

Moles = Molarity x volume in L

= 0.1 X 100/1000

= 0.01 moles

Now moles = Mass in g/molar mass

Mass in g = moles x molar mass

= 0.01 x 111

= 1.11 g

So the mass of CaCl_{2} in 100ml of the solution is 1.11 g

The term is quite useful in all fields of chemistry. In pharmaceutical industry measuring the concentration in molarity is the backbone of all preparations.Change in molarity can cause aquatic life cease to exist.

Another term, Molality or molal concentration can be used.

The** difference between molarity and molality** is that solvent is taken in kilograms in molality, while we take the volume of the solvent in molar calculations. Sometimes, it is difficult to calculate the mass of a solvent, say water or ethanol. So, to avoid such discrepancies, Molar concentrations are widely used.

**Molar Concentration**

For
one mole of solute dissolved in one liter of solution, molarity, M = 1.
Such a solution is called as 1 molar solution. A solution containing
two moles of solute in one liter is 2M, 2 molar solution, and so on.

Molarity can also be expressed in terms of **Avogadro's number.**

Avogadro's number is the number of molecules present in one mole of any substance.The number has been determined to be 6.022 x 10^{23}.

So the molarity in terms of no. of molecules is N / 6.022 x 10^{23} X V.

where N is number of molecules of solute,V is volume of solvent.

Molarity =$\frac{ No. of molecules }{(N \times V)}$

=$\frac{8.283 \times 10^23}{6.022 \times 10^23 \times 2}$

= 0.68 M

Molarity = No. of molecules /N X V

1.7 = No. of molecules / 6.022 x 10^{23} x 1.43

No. of molecules=1.7 X 6.022 x 10^{23} x 1.43

=14.63 Xx 10^{23}

Some of the problems based on molarity from density is given below.

First convert mass of water to volume.

**Volume =** $\frac{Mass}{density}$

=$\frac{100}{1}$=100ml

**Molarity =** $\frac{Moles}{volume}$

Moles =${11.8}{58.4}$ = 0.2

Molarity =${ 0.2}{0.1}$

= 2

Molarity is defined for a specific temperature because increase in temperature increases the volume.

Molarity of HCl = 0.540 M

Calculating the volume 155ml x $\frac{1\ liter}{1000\ ml}$ = 0.155 liters.

Number of moles of HCl = Molarity x Volume in liters

= 0.540 M x 0.155 liters

= 0.0837 moles of HCl.

Since we need to calculate the weight in grams of HCl present, we can convert the moles to grams.

Molar mass of HCl = 36.5 grams/mol.

0.0837 moles of HCl x $\frac{36.5\ grams}{1\ mole\ of\ HCl}$ =**3.06 grams Of HCl.**

So, weight in grams of HCl can be calculated from its Molarity value.

Calculating the volume 155ml x $\frac{1\ liter}{1000\ ml}$ = 0.155 liters.

Number of moles of HCl = Molarity x Volume in liters

= 0.540 M x 0.155 liters

= 0.0837 moles of HCl.

Since we need to calculate the weight in grams of HCl present, we can convert the moles to grams.

Molar mass of HCl = 36.5 grams/mol.

0.0837 moles of HCl x $\frac{36.5\ grams}{1\ mole\ of\ HCl}$ =

So, weight in grams of HCl can be calculated from its Molarity value.

Some of the problems based on molarity is given below.

Moles = Mass in g / Molar mass

Moles of NaCl =116.8 / 58.4

Moles=2

**Step II:Find molarity**** **

Molarity =$\frac{moles of solute NaCl }{ Volume of solution in L}$

= $\frac{2 }{1}$

Molarity = 2 M

Molar mass of Silver nitrate = 108 g/mole.

So, Molarity of solution = 0.04M.

n(AgNO_{3}) = 1.08 grams of AgNO_{3} x $\frac{1\ mole\ of\ AgNO_3}{108\ grams\ AgNO_3}$ = **0.01 mole. **

Number of moles in 1 dm^{3} = 0.01 mole x $\frac{1000}{250}$^{} = 0.04 moles/dm^{3}

So, Molarity of solution = 0.04M.

Molecular weight of Sodium Hydroxide, NaOH = 40 grams/mol

We know that

**Molarity =**

$\frac{number\ of\ moles}{decimeter\ cube\ of\ the\ solution\ or\ number\ of\ moles\ per\ liter}$

we need two things here

1. Moles of NaOH and

2. Volume in liters.

Calculating the moles

Calculating the volume: 540ml x $\frac{1\ liter}{1000\ ml}$ = 0.540 liters.

Molarity of NaOH = n/v = $\frac{1.885\ moles}{0.540\ liters}$ = 3.49 moles/Liters or**3. 5M**

We know that

we need two things here

1. Moles of NaOH and

2. Volume in liters.

Calculating the moles

75.5 grams of NaOH x $\frac{1\ mole}{40\ grams\ of\ NaOH}$ = 1.885 moles

Calculating the volume: 540ml x $\frac{1\ liter}{1000\ ml}$ = 0.540 liters.

Molarity of NaOH = n/v = $\frac{1.885\ moles}{0.540\ liters}$ = 3.49 moles/Liters or

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