Chemistry - Class Twelve

Solutions

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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 Solutions class 12 chemistry - NCERT In Text Solution11

Solutions class 12 chemistry - NCERT In Text Solution12

Now, we know that, Molarity Solutions class 12 chemistry - NCERT In Text Solution13

Solutions class 12 chemistry - NCERT In Text Solution14

Thus, molarity of given solute = 0.24 M

(b) Given, 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Thus,

Solutions class 12 chemistry - NCERT In Text Solution15

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 Solutions class 12 chemistry - NCERT In Text Solution16

Now, we know that, Molarity Solutions class 12 chemistry - NCERT In Text Solution17

Solutions class 12 chemistry - NCERT In Text Solution18

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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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