1.

The volume of H_(2)liberated at 273K and 4 atm pressure by the passage of 1 faraday of electricitythrough HCl solution is

Answer»

`4.48 dm^(3)`
`5.6 dm^(3)`
`2.8 dm^(3)`
`112 dm^(3)`

Solution :1F=1 mole of `E^(-)-=1` mole of H ATOM
`-=1//2" mole of "H_2 = 11.2 dm^3 of H_2`
Now,`(P_1V_1)/(T_1) =(P_0V_0)/(T_0)`
`therefore (4xxV_1)/(273)=(1xx11.2)/(273) therefore (4V_1)/(1) =(11.2)/(1)`
`therefore V_1 =2.8 dm^3`


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