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Thursday 17 March 2022

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.



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