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Two solution of H(2)SO(4) of molarities ...

Two solution of `H_(2)SO_(4)` of molarities x and y are mixed in the ratio of `V_(1) mL : V_(2) mL` to form a solution of molarity `M_(1)`. If they are mixed in the ratio of `V_(2) mL : V_(1) mL`, they form a solution of molarity `M_(2)`. Given `V_(1)//V_(2) = (x)/(y) gt 1 and (M_(1))/(M_(2)) = (5)/(4)`, then `x : y` is

A

`2:1`

B

`4:1`

C

`1:2`

D

`3:1`

Text Solution

Verified by Experts

Molarity of the mixture can be calculated as:
`M_(1)V_(1)+M_(2)V_(2)=M_(R)(V_(1)+V_(2))`
where `M_(R)=` resultant solution
`V_(1)xx x+V_(2) xx y=M_(1)(v_(1)+v_(2))`…………..i
`V_(2)xx x+V_(1)xxy=M_(2)(V_(1)+V_(2))` ………ii
Dividing eq (i) by eq. (ii) , we get
`(V_(1)x+V_(2)y)/(V_(2)x+V_(1)y)=(M_(1))/(M_(2))`
Substituting `(M_(1))/(M_(2))=5/4` and `(V_(1))/(V_(2))=x/y` we can calculate x:y.
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