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100 ml, 0.1 M K(2)SO(4) is mixed with 10...

100 ml, 0.1 M `K_(2)SO_(4)` is mixed with 100 ml, 0.1M `Al_(2)(SO_(4))_(3)` solution. The resultant solution is

A

`0.1NK^(+)` ions

B

`0.4NK^(+)` ions

C

`0.2MSO_(4)^(-2)` ions

D

`0.1MAl^(+3)` ions

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`[K^(+)]=(100xx0.1xx2)/(200)=0.1M=0.1N`
`[SO_(4)^(2-)]=(100xx0.1xx1+100xx0.1xx3)/(200)`
`=0.2M=0.4N`
`[Al^(+3)]=(100xx0.1xx2)/(200)=0.1M=0.3N`
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