Solution of NCERT In Text Question-2.3
Question - 2.3 - Calculate the molarity of each of the following solutions:
(a) 30 g of Co(NO 3) 2. 6H2 O in 4.3 L of solution
(b) 30 mL of 0.5 M H 2SO4 diluted to 500 mL.
Solution:
(a) Given,
Mass of solute (W B) = 30 g
Molar mass of given solute Co(NO 3) 2.6H 2O = 58.7 + 2[14 + (16 x 3)] + 6 ( 2 + 16)
= 58.7 + (2 x 62) + (6 x 18)
= 58.7 + 124 + 128 g mol -1
= 290.7 g mol -1
Now, Number of moles of Co(NO 3)2 .6H 2O
Now, we know that, Molarity
Thus, molarity of given solute = 0.24 M
(b) Given, 30 mL of 0.5 M H2SO4 diluted to 500 mL.
Thus,
Thus, required molarity = 0.3 M
Alternate method:
Number of moles present in 1000 ml of 0.5 M H 2SO4 solution = 0.5 mol
Therefore, number of moles present in 1 ml of solution = 1 / 1000 mL
Therefore, number of moles present in 30 mL of solution
Now, we know that, Molarity
