`0.5 M FeSO_(4)` विलयन के 35 mL को ऑक्सीकृत करने के लिए आवश्यक 0.1 M `K_(2) Cr_(2)O_(7)` का आयतन है :

Options

A. 29.2 mL
B. 175 mL
C. 185 mL
D. 145 mL

Answer

A

Explanation

`Cr_(2)O_(7)^(2-) +14H^(+) +6e^(-) +6e^(-) to 2Cr^(3+) +7H_(2)O`
`a_(1)=6`
`a_(2)=1`
`a_(1)=K_(2)Cr_(2)O_(7)` का n-गुणांक
`a_(2)= FeSO_(4)` का n-गुणांक
`Fe^(2+) to Fe^(3+) + e^(-)`
`underset(a_(1)=6) underset((K_(2)Cr_(2)O_(7))) (a_(1)M_(1)V_(1)) =underset(a_(2)=1) underset((FeSO_(4))) (a_(2)M_(2)V_(2))`
`M_(1)=0.1 M`
`M_(2)=0.5M`
`V_(1)=`?
`V_(2)=35mL`
इसलिए, `V_(1)=(1xx0.5xx35mL)/(6xx0.1)`
=`29.2mL`.