Chemistry For Class IX (New Book ) - Chapter No. 6 - Solutions - Numericals

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Chapter No.6- Solutions
Numericals
Text Book Exercise

More Numericals
Examples From Text Book

Example 6.2: How would you prepare 500 cm³ of 0.20 M NaOH (aq) from a stock solution of 1.5 M NaOH?
Solution:
Data:

• M₁ = 1.5 M NaOH
• M₂ = 0.2 M NaOH
• V₂ = 500 cm³
• V₁ = ?

Calculation:
Formula:

Concentrated Soluiotn = Dilute Solution

M₁ V₁ = M₂ V₂

1.5 x V₁ = 0.20 x 500

= 66.67 cm³

Ans: Take 66.67 cm3 of concentrated NaOH and placed in a measuring flask and dilute it by adding water up to the mark. It will become 0.2M solution of NaOH.



Example 6.5: 20 gram of salt is dissolved in 500 cm³ of a solution. Calculate the molarity of that solution.
Solution:
Data:

• Mass of solute = 20 g
• Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
• Volume of solution = 500 cm³
• Molarity (M) = ?

Ans: The molarity of that solution is 0.683 mole / dm³.





Example 6.7: A sample of sulphuric acid has the molarity 20 M. How many cm³ of solution should you use to prepare 500 cm³ of 0.5 M H₂SO₄?
Solution:
Data:

• M₁ = Molarity of given H₂SO₄ = 20 M
• M₂ = Molarity of required H₂SO₄ = 0.5 M
• V₁ = volume of concentrated solution needs to be diluted =? 1
• V₂ = Volume required in H₂SO₄ on = 500 cm³

Formula:

M₁ V₁ = M₂ V₂

Calculation:

V₁ = 12.5 cm³

Ans: 12.5 cm³ of 20 M is used to make 500 cm³ aqueous solution to form 0.5 M H₂SO₄.