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

Q3.

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

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Important Questions & Answers

What is the correct answer to: 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&...

How do you solve 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. step by step?

Step-by-step explanation:
• Molarity is given by:

• Molarity =  moles of solute / Volume of solution in litre

• (a) Molar mass of Co(NO3)2

Why is this answer important for exams?

This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.

Which NCERT concept is used in this question?

This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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