Friday, 3 May 2013

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.


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.
Solution
(a)
Given that mass of Co (NO3)2.6H2O  = 30 g
Volume of solution = 4.3 liter
Molar mass of Co(NO3)2.6H2O = 59 + 2(14 + 3 × 16) + 6 (16 +1 × 2) = 291 g mol−1
Use the formula 
Moles of Co(NO3)2.6H2O = 30/291 = 0.103 mol
Use the formula 
Molarity of Co(NO3)2.6H2O = 0.103/4.3= 0.023 M

(b)
Given that
Volume of H2SO4 (V1) = 30 ml)
Diluted volume of H2SO4 (V2) = 500 ml

Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
Initial molarity M1= 0.5 M 
We have to find M2
Use the formula
M2V2 = M1V1
plug the values we get
M2  × 500  = 0.5 × 30
M2 = 15/ 500 = 0.03 M
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31 comments:

  1. Unknown15 May 2016 at 01:20

    Fuck this answer it doesn't have proper answer

    ReplyDelete
  2. Unknown15 July 2017 at 12:00

    Thanks for the answer. It is correct and well explained.

    ReplyDelete
  3. Unknown27 July 2017 at 11:20

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  4. Unknown13 March 2018 at 09:07

    Thanks

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  5. Unknown27 April 2018 at 22:01

    It is a correct answer nd well explained..

    ReplyDelete
  6. Sudip29 June 2018 at 19:42

    Nice answe.ur solution is always good

    ReplyDelete
  7. Unknown30 June 2018 at 21:41

    Thanks bro my answer is also same but in pradeep refresher and on some site answer is wrong

    ReplyDelete
  8. Unknown9 August 2018 at 02:13

    What the hell does 30 g of 0.5 M H2SO4 diluted in 500 ml mean?

    ReplyDelete
    Replies
    1. Dhriti6 April 2019 at 04:06

      Its 30ml 0.5M means molarity of solution is initially 0.5 and its volume is 30ml

      Delete
  9. PRABHUDEEN24 September 2018 at 02:39

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  10. Unknown31 December 2018 at 22:54

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  11. Unknown11 January 2019 at 19:45

    Thanks for solution.

    ReplyDelete
  12. Manisha gupta20 January 2019 at 16:52

    Thnks

    ReplyDelete
  13. Unknown26 February 2019 at 04:36

    Thnk u for solution

    ReplyDelete
  14. Unknown12 March 2019 at 08:43

    thanks a lot for solution

    ReplyDelete
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  16. yash19 March 2019 at 17:30

    Answer is very helpful..tq sir

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  17. Unknown5 April 2019 at 01:32

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  18. Unknown5 April 2019 at 01:32

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  19. Unknown5 April 2019 at 07:34

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  20. Unknown7 April 2019 at 22:28

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  21. Unknown6 May 2019 at 12:12

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  22. Unknown19 May 2019 at 10:08

    It is helpful

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  23. Unknown15 June 2019 at 11:41

    Thanks sir it is helpful

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  24. Unknown4 March 2020 at 07:07

    Thanks sir

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  25. Unknown4 March 2020 at 07:08

    Thanks sir

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