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To 50 mL of 0.5 M H(2)SO4 75 mL of 0.25 ...

To 50 mL of 0.5 M `H_(2)SO_4` 75 mL of 0.25 M `H_2SO_4` is added. What is the concentration of the final solution if its volume is 125 mL?

Text Solution

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No. of moles in 0.05 litre of `H_(2)SO_(4) = M xx V` (litre)
`=0.5 xx 0.05 = 0.0025`
No. of moles in 0.0075 litre of `H_(2)SO_(4)` added `=0.25 xx 0.075`
`=0.01875`
Total no. of moles in 0.125 litre of `H_(2)SO_(4) = 0.025 + 0.01875`
=0.04375
`therefore` Molarity of `H_(2)SO_(4) = (0.04375)/(0.0125)= 0.35 M`
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