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 acidulated water, then the volume of H_(2) liberated at STP will be [given : at.wt. of Cu = 64].

Answer»

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

Solution :`("wt. of Cu deposited")/("wt. of " H_(2) " produced")=("EQ. wt. of Cu")/("eq. wt. of H")=(64//2)/(1)`
or,`(0.16)/("wt. of " H_(2))=(32)/(1)`
or, wt. of `H_(2) =(0.16)/(32)=5XX10^(-3)g`.
`therefore`VOLUME of `H_(2)`liberated at STP
`=(22400)/(2)xx5xx10^(-3)` cc = 56 cc.


Discussion

No Comment Found

Related InterviewSolutions