Question

how prepare aqueous solution :
355ml of 0.0956M LiNo3 from 0.244M LiNo3 (MLiNo3=69g/mol)

Answer 1

This Problem can be solved by using Dilution Law I.e.

Where ,

M₁ = Intial concentration = 0.244 M

V₁ = intial volume = Unknown (we have to find this) = let this be 'y'

M₂ = final concentration = 0.0956 M

V₂ = final volume = 355 mL

Applying the dilution law,

0.244 M * y mL = 0.0956 M * 355 mL

y = (0.0956 * 355 ) / 0.244 mL

y = 139 mL

so the intial Volume V₁ = 139 mL

Now, How to prepare?

Take 139 mL of 0.244 M LiNO₃ and add (355 - 139) mL of water

i.e. Take 139 mL of 0.244 M LiNO₃ and add 216 mL of water

Therefore the resultant solution has the concentration 0.0956 M

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