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on boiling 1litre n/5 hcl the volume of aqueous solution decrease to 250ml. during this if 3.65 from of hcl is removed from the solution what will be the concentration of resulting solution.
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on boiling 1litre n/5 hcl the volume of aqueous solution decrease to 2...
Given:

Initial volume of HCl solution (Vi) = 1 L

Initial concentration of HCl solution (Ci) = n/5

Final volume of HCl solution (Vf) = 250 mL = 0.25 L

Amount of HCl removed from the solution (n) = 3.65 g

Calculations:

Initial amount of HCl present in the solution = Ci x Vi = (n/5) x 1 = n/5 g

Final amount of HCl present in the solution = Initial amount of HCl - Amount of HCl removed = (n/5) - 3.65 g

Final concentration of HCl solution (Cf) = Final amount of HCl / Final volume of HCl solution = [(n/5) - 3.65 g] / 0.25 L

Answer:

The concentration of the resulting HCl solution is [(n/5) - 3.65 g] / 0.25 L.
Community Answer
on boiling 1litre n/5 hcl the volume of aqueous solution decrease to 2...
N/2.5
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on boiling 1litre n/5 hcl the volume of aqueous solution decrease to 250ml. during this if 3.65 from of hcl is removed from the solution what will be the concentration of resulting solution.
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