2. Solutions Class 12 Chemistry Chapter 2 Solutions, Class 12 Chemistry | Page 12

Molarity o Molarity ( M ) can be defined as the ratio of number of moles of solute to the volume of solution in liter . o
Molarity o Molarity ( M ) can be defined as the ratio of number of moles of solute to the volume of solution in liter . o
o
o
o
It can also be defined as the number of moles of solute present in unit litre or cubic decimeter of solution .
It is function of temperature due to the dependence of volume on temperature whereas Mass %, ppm , mole fraction and molality are independent of temperature because mass does not depend on temperature .
Mathematically , Molarity = Moles of Solute / Volume of Solution in Litre
For instance , molarity of a solution containing 0.25 mol of NaOH dissolved in one litre or one cubic decimeter is 0.25 M or 0.25 mol L – 1 .
Problem : Calculate the molarity of each of the following solutions : ( a ) 30 g of Co ( NO3 ) 2 . 6H2O in 4.3 L of solution ( b ) 30 mL of 0.5 M H2SO4 diluted to 500 mL . Solution : Molarity = Moles of Solute / Volume of Solution in Litre ( a ) Molar mass of Co ( NO3 ) 2.6H2O ( solute )
= 59 + 2 ( 14 + 3 × 16 ) + 6 × 18 = 291 g mol - 1 ∴Moles of Co ( NO3 ) 2.6H2O = 30 / 291 mol = 0.103 mol Therefore , molarity = 0.103 mol / 4.3 L = 0.023 M ( b ) Number of moles present in 1000 mL ( Solvent ) of 0.5 M H2SO4 ( Solute ) = 0.5 mol
∴ Number of moles present in 30 mL of 0.5 M H2SO4 = ( 0.5 X 30 ) / 1000 mol = 0.015 mol
Therefore , molarity = 0.015 mol / 0.5 L = 0.03 M