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. $56\, cm^3$
• B. $550\, cm^3$
• C. $5.6\, cm^3$
• D. $11.2 \,cm^3$

Answer 1

Answer: (C)

$5.6\, cm^3$

Answer 2

\frac{\text{Mass of Agdeposited}}{\text{Mass ofO_{2}evolved}} = \frac{\text{E mass of} Ag}{\text{E mass of} O_{2} } $\frac{0.108}{m\left(O_{2}\right)} = \frac{108}{8}$ $m\left(O_{2}\right) =\frac{ 8\times 0.108}{108}$ $= 8 \times 10^{-3}$ $32 \,g\, O_{2} = 22400 \,cm^{3}$ at N.T.P. $\therefore 8\times 10^{-3} g$ of $O_{2}$ $= \frac{22400}{32}\times8 \times 10^{-3}cm^{3}$ at N .T.P. $= 5.6\, cm^{3} N.T.P$