1.

x g of KHC_(2)O_(4) requires 100 ml of 0.02 M–K MnO_(4) in acidic medium. In another experiment, y g of KHC_(2)O_(4) requires 100 ml of 0.05 M-Ca(OH)_(2) . The ratio of x and y is

Answer»

`1:1`
`1:2`
`2:1`
`5:4`

Solution :`n_(eq)KHC_(2)O_(4) = n_(eq) KMnO_(4)` or
`x/M xx 2 = (100 xx 0.2)/(1000) xx 5`
`n_(eq)KHC_(2)O_(4) = n_(eq)CA(OH)_(2)`
or. `y/M xx 1 = (100 xx 0.5)/1000 xx 2`
`:. x/y = 1/2`.


Discussion

No Comment Found

Related InterviewSolutions