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

31 comments:

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

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

    ReplyDelete
    Replies
    1. Its 30ml 0.5M means molarity of solution is initially 0.5 and its volume is 30ml

      Delete
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