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)

$\frac{\text{Mass of $Ag$ deposited}}{\text{Mass of $O_{2}$ evolved}} = \frac{\text{Eq. mass of} Ag}{\text{Eq. 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.
$= 5.6\, cm^{3} N.T.P$