It is quite widely used unit and it is denoted by letter M
It is the no. of moles of solute present in per litre solution. Unit = Moles per lit.
Molarity Formula is given as
If Volume = 1 litre
M = n solute
Unit of Molarity is Moles L-1
In terms of weight, molarity formula of the substance can be expressed as:
Molarity formula with Density is given as
How to calculate the Molarity of a solution
1. First ,we need to calculate the Molar Mass
of the solute. This can be find using the molar mass of the elements
Let's take a example of NaOH
Molar mass of NaOH = Molar mass of Na + Molar mass Of O + Molar mass of H
=23+16+1=40
2. We calculate the Moles of the solute by the formula
$n=\frac {\text{Mass(g)}}{\text{Molar Mass}}$
3. Find the Volume of Solution in Litre
We can use below table for our convenience
1 cm3
1mL
1000 mm3
1 Litre
1000mL
1000 cm3
1 m3
106 cm3
1000 L
1 dm3
1000 cm3
1 L
4. Now Molarity can be simply calculated using the Formula of Molarity
$M= \frac {\text{No Of Moles of Solute}}{\text {Volume of Solution in (L)}}$
Solved Example of Molarity Formula
Question 1
Calculate the molarity of $NaOH$ in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. Solution
As per Molarity Formula
$M = \frac {W}{M} \times \frac {1000}{\text{V ml}}$
= (4 g / 40 g ) × (1000/250) =0.1 mol/0.250 L
= 0.4 mol/L
= 0.4 M
Question 2
If 2 gm $NaOH$ is dissolute in water to make solution upto 250 cc. Give molarity of the solution: Solution
NaOH = W = 2 g
Molar Mass of NaOH = 23 + 16 + 1 = 40
M = W/M × 1000/(V ml) = 2/40 × 1000/250=1/5 M
Question 3
Give the mass of NaOH dissolute to make solution of 500 cm3, 1/2 M molar. Solution
M = W/M × 1000/(V ml)
Molar Mass of NaOH = 23 + 16 + 1 = 40
M=1/2 and V=500 ml
Therefore,
1/2=W/40 ×1000/500
W=20/2=10gm
Question 4
20% (w/W) NaOH solution (density of solution= 1.2 g/cc). Calculate the Molarity Solution
20% (w/M) means 20 gm in 100gm of solution
Now Density of solution =1.2g/cc
So Volume of Solution =100/1.2 cc
Now
M = W/M × 1000/(V ml)
M= 20/40 × 1000/100 × 1.2 = 6 M Question 5
Give the volume of the solution if 10 gm NaOH, 1/20 M NaOH sol. is present. Solution
M = W/M × 1000/(V ml)
1/20=10/40 × 1000/(V ml)
4/20= 1000/(V ml)
V = 5000 ml
= 5L Question 6
What is the amount of solute present in M/40,100 ml Na OH sol. Solution
M = W/M × 1000/(V ml)
1/40=W/40 ×1000/100
4/40 = W
0.1 gm = W Question 7
Calculate the morality of 2% (m/v) glucose sol.) Solution
2 gm Glucose in 100 ml sol.
Now Molar mass of glucose(C6 H12 O6) = 12×6+1 ×12+6
= 72 + 12 + 96=180 gm
M = W/M × 1000/(V ml)
M = 2/180 ×1000/100
M = 1/9
Molarity equation
To calculate the volume of a definite solution required to prepare solution of other molarity, the following equation is used:
M1 V1 = M2 V2
where M1= initial molarity
M2= molarity of the new solution
V1= initial volume
V2= volume of the new solution
Suppose we have mix three solution of same solute of Molarity M1 ,M2 ,M3 and Volume V1,V2 ,V3
Molarity of the resulting solution is given by
$M_R= \frac {M_1V_1 +M_2V_2 + M_3V_3}{V_1 + V_2 + V_3}$
Solved Example of Molarity Equations
Question 1
If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
(i) 1.5 M
(ii) 1.66 M
(iii) 0.017 M
(iv) 1.59 M Solution
M1 V1 = M2 V2
500 X 5=1500 M2
M2 =1.66 M Question 2
If 4 g of NaOH dissolves in 36 g of H2O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific
gravity of solution is 1g mL–1). Solution
Mass of NaOH = 4 g
Number of moles of NaOH =4/40
= 0.1 mol
Mass of H2O = 36 g
Number of moles of H2O= 36/18 =2 mol
Mole fraction of water
=Number of moles of H2O, /(No. of moles of water + No. of moles of NaOH)
=2/(2+.1) = 0.95
Mole fraction of NaOH =Number of moles of NaOH/ (No. of moles of NaOH + No. of moles of water)
= .1 /(2+.1) = 0.047
Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g
Volume of solution = 40 × 1 = 40 mL
(Since specific gravity of solution is = 1 g mL–1)
Molarity of solution
=Number of moles of solute/ Volume of solution in litre
=0.1 mol NaOH /.04 = 2.5 M Question 3
How do I prepare 0.2 M HCL? Solution
Now As per Molarity Formula
For 0.2M , we need 0.2 moles of HCl in 1 Litre Solution
Now Molecular mass of HCl=34.46 g/mol
1 mole of HCl contain $1 \times 34.46$ g HCl (no.of moles= given mass of solute/ molecular mass of solute)
0.2 moles of HCl contain $0.2 \times 34.46=6.892$g HCl
Therefore 6.892 g of HCl in a litre of solution gives 0.2 molar HCl solution
Relation Between Mole Fraction and Molarity
Let χA = Mole fraction of solvent
χB = Mole Fraction of solute
nA = Number of moles of solvent
nB = Number of moles of solute
M = molarity of solution
d = Density of solution
MA = Molar mass of solvent
MB = Molar mass of solute
Now Mass of solution = nA MA + nBMB
Volume of solution = Mass of solution/density of solution
So, Volume of solution = (nA MA + nBMB)/d
The density of a solution is in g/mL
Molarity = Number of moles of Solute/ Volume of the solution
=nB × d × 1000/(nA MA + nBMB)
Dividing Numerator and denominator by nA+nB
= χB × d × 1000/(χA MA + χBMB)
