Class 12 Chemistry - Chapter Solutions NCERT Solutions | Calculate the molarity of each of the fo

Welcome to the NCERT Solutions for Class 12th Chemistry - Chapter Solutions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 3: calculate the molarity of each of the following so....
Question 3

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.

Answer

Molarity is given by:

Molarity =  moles of solute / Volume of solution in litre

(a) Molar mass of Co(NO3)2.6H2O

= 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

= 0.024 M

 

(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 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

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