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calistafernand
calistafernand calistafernand
  • 17-06-2016
  • Chemistry
Answered

What volume of 0.25 mol/L solution of lead(II)nitrate would be required to form 500mL of a 0.15 mol/L solution of lead(II)nitrate. Show calculations

Verified Answer :

taskmasters
taskmasters taskmasters
  • 25-06-2016

To solve this we use the equation,

 M1V1 = M2V2

 where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.

0.25 M x V1 = 0.15 M x .500 L

V1 = 0.3 L or 300 mL

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