Same amount of electric current is passed through solutions of `AgNO_(3)` and HCl. If 1.08 g of silver is obtained in the first case, the amount of hydrogen liberated at S.T .P. in the second case is:A. `112cm^(3)`B. `22400cm^(3)`C. `224cm^(3)`D. `1.008g`.

Same amount of electric current is passed through solutions of `AgNO_(

1.	Same amount of electric current is passed through solutions of `AgNO_(3)` and HCl. If 1.08 g of silver is obtained in the first case, the amount of hydrogen liberated at S.T .P. in the second case is:A. `112cm^(3)`B. `22400cm^(3)`C. `224cm^(3)`D. `1.008g`.
Answer» Correct Answer - A `("Mass of Ag")/("Mass of " H_(2))=("Eq. mass of Ag")/("Eq. mass of " H_(2))` `(1.08)/("mass of" H_(2))=(108)/(1)` Mass of `H_(2)=( 1.08)/(108)=10^(-2)g` 2g of `H_(2)` at S.T.P. occupy 22400 `cm^(3)` `therefore10^(-2)g` of `H_(2)` at STP occupy `=(22 400)/(2)xx10^(-2)10^(-2)cm^(3)=112cm^(3)`

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Same amount of electric current is passed through solutions of `AgNO_(3)` and HCl. If 1.08 g of silver is obtained in the first case, the amount of hydrogen liberated at S.T .P. in the second case is:A. `112cm^(3)`B. `22400cm^(3)`C. `224cm^(3)`D. `1.008g`.

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