Same amount of electric current is passed through solutions of AgNO3 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

Question

Same amount of electric current is passed through solutions of AgNO3 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) 112 cm3 (B) 22400 cm3 (C) 224 cm3 (D) 1.008 g

Tardigrade Admin · Accepted Answer

( textMass of Ag/ textMass of H2) = ( textEq. mass of Ag/ textEq. mass of H2) (1.80/ textMass of H2) = (108/1) Mass of H2 = (1.80/108) = 10-2 g 2 g of H2 at S.T.P. occupy 22400 cm3 ∴ 10-2 g of H2 at S.T.P. occupy = (22400/2)× 10-2 cm3 = 112 cm3

Q. Same amount of electric current is passed through solutions of $A g N O_{3}$ 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

112 $c m^{3}$

22400 $c m^{3}$

224 $c m^{3}$

1.008 g

Q. Same amount of electric current is passed through solutions of AgNO3​ 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

112 cm3

22400 cm3

224 cm3

1.008 g

Mass ofH2​Mass ofAg​=Eq. mass ofH2​Eq. mass ofAg​ Mass ofH2​1.80​=1108​ Mass of H2​=1081.80​=10−2g 2g of H2​ at S.T.P. occupy 22400cm3 ∴10−2g of H2​ at S.T.P. occupy =222400​×10−2cm3 =112cm3

Q. Same amount of electric current is passed through solutions of $A g N O_{3}$ 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

112 $c m^{3}$

22400 $c m^{3}$

224 $c m^{3}$