An electric current is passed through silver voltameter connected to a water voltmeter. The cathode of the silver voltameter is 0.108 g moreat the end of the electrolysis. The volume ofoxygen evolved at STP is :

Question

An electric current is passed through silver voltameter connected to a water voltmeter. The cathode of the silver voltameter is 0.108 g moreat the end of the electrolysis. The volume ofoxygen evolved at STP is : (A) 56 cm3 (B) 550 cm3 (C) 5.6 cm3 (D) 11.2 cm3

Tardigrade Admin · Accepted Answer

( textMass of Ag deposited/ textMass of O2 evolved) = ( textEq. mass of Ag/ textEq. mass of O2 ) (0.108/m(O2)) = (108/8) m(O2) =( 8× 0.108/108) = 8 × 10-3 32 g O2 = 22400 cm3 at N.T.P. ∴ 8× 10-3 g of O2 = (22400/32)×8 × 10-3cm3 at N .T.P. = 5.6 cm3 N.T.P

Q. An electric current is passed through silver voltameter connected to a water voltmeter. The cathode of the silver voltameter is 0.108 g moreat the end of the electrolysis. The volume ofoxygen evolved at STP is :

$56 c m^{3}$

$550 c m^{3}$

$5.6 c m^{3}$

$11.2 c m^{3}$

Q. An electric current is passed through silver voltameter connected to a water voltmeter. The cathode of the silver voltameter is 0.108 g moreat the end of the electrolysis. The volume ofoxygen evolved at STP is :

56cm3

550cm3

5.6cm3

11.2cm3

Mass of O2​ evolvedMass of Ag deposited​=Eq. mass ofO2​Eq. mass ofAg​ m(O2​)0.108​=8108​ m(O2​)=1088×0.108​ =8×10−3 32gO2​=22400cm3 at N.T.P. ∴8×10−3g of O2​ =3222400​×8×10−3cm3 at N .T.P. =5.6cm3N.T.P

$56 c m^{3}$

$550 c m^{3}$

$5.6 c m^{3}$

$11.2 c m^{3}$