1.

When a quantity of electricity is passed through CuSO_(4) solution 0.16 g of copper gets deposited if the same quantity of electricity is passed through acidullated water, then the volume of H_(2) gasliberated at S.T.P. will be (Given at wt. of Cu=64)

Answer»

`4.0cm^(3)`
`56cm^(3)`
`604cm^(3)`
`8.0cm^(3)`

SOLUTION :`("Mass of CU deposited")/("Volume of" H_("liberated"))=("Eq. WT of Cu")/("Eq. wt. of H")`
`(0.18)/("Mass of" H_(2) "liberated")=(64//2)/(1)`
or mass of `H_(2)` liberated `=(0.16)/(32)g=5xx10^(-3)g`
`therefore` volume of `H_(2)` liberated at STP
`=(22400)/(2)xx5xx10^(-3)cm^( 3)=56cm ^(3)`


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