# Solutions

## 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

Topics covered in this chapter of Solutions

1. Solutions

2. Types of Solutions

3. Solutions - Concentration

4. Solution of NCERT In Text Questions

Class 12 Biology

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