An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed $0.108 g$ more at the end of the electrolysis. The volume of $O_{2}$ evolved at STP :

56 {c m}^{3}

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550 {c m}^{3}

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5.6 {c m}^{3}

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11.2 {c m}^{3}

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Solution

The correct option is C $5.6 {c m}^{3}$
$\frac{W t . o f s i l v e r d e p o s i t e d}{W t . o f O_{2} p r o d u c e d} = \frac{E q . w t o f A g}{E q . w t o f O_{2}}$

$\frac{0.108}{W t . o f O_{2}} = \frac{108}{8}$

$∴$ Wt. of $O_{2}$ produced $= \frac{0.108 \times 8}{108} = 0.008 g$

Now, $32 g$ of $O_{2}$ produces $= 22400 m l$ of $O_{2}$

$∴$ $0.008 g$ of $O_{2}$ produces $\frac{22400}{32} \times 0.008 = 5.6 {c m}^{3}$ of $O_{2}$

Hence, option C is correct.

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An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed 0.108g more at the end of the electrolysis. The volume of O2 evolved at STP :

The correct option is C 5.6 cm3Wt.ofsilverdepositedWt.ofO2produced=Eq.wtofAgEq.wtofO20.108Wt.ofO2=1088∴ Wt. of O2 produced =0.108×8108=0.008gNow, 32g of O2 produces =22400 ml of O2∴ 0.008g of O2 produces 2240032×0.008=5.6 cm3 of O2Hence, option C is correct.

An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed $0.108 g$ more at the end of the electrolysis. The volume of $O_{2}$ evolved at STP :