10 gram of hydrogen is liberated by the electrolysis of NaOH solution at the cost of Rs.20. The cost of current to liberate 10gram of oxygen under similar conditions is in :

Rs.1.50

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Rs.20

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Rs 2.50

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Rs. 25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Rs 2.50
$10 g H_{2} \Rightarrow \frac{10}{2} = 5 m o l e s o f H_{2}$
$2 H^{+} + 2 e^{-} \to H_{2}$
1 mole $H_{2}$ requires 2 moles $e^{-}$
$∴ m o l e o f e^{-} = 5 \times 2 = 10 m o l e e^{-}$
$10 m o l e e^{-} c h a r g e c o s t R s 20$
$∴ 1 m o l e c o s t s = \frac{R s 20}{10} = R s 2$
$10 g O_{2} = \frac{10}{32} m o l e s o f O_{2}$
$4 O H^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}$
1 mole $O H^{-}$ liberates 4 mole of $e^{-}$
$∴ \frac{10}{32} m o l e s l i b e r a t e = \frac{10 \times 4}{32} = \frac{5}{4} m o l e e^{-}$
$∴ p r i c e o f e l e c t r i c i t y = \frac{5}{4} \times 2 = R s 2.5$

Suggest Corrections

Similar questions

N a O H, 2.8 l i t r e

N a O H, 2.8

What volume of $H_{2}$ is liberated under similar conditions which released $2.24 d m^{3} O_{2}$ by electrolysis of water?

O_{2}

2.24 d m^{3}

S T P

22.4 L

Join BYJU'S Learning Program