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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 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 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.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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.
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
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 3: Calculate the molarity of each of the following solutions: (a)30 g of Co(NO3)2. 6H2O in 4.3 L of ....
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