On passing a current through a molten aluminium chloride for some time produced 11.28lit of $C l_{2}$ at NTP at anode. The quantity of aluminium deposited at cathode is:

27grams

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

18gram

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

9gram

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

36 gram

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

Open in App

Solution

The correct option is D 9gram
$11.2 l o f c l_{2} a t N T P$
$\Rightarrow 0.5 m o l e o f c l_{2}$
$c l_{2}$ gives 2 moles of electer
$2 c l^{-} + 2 e^{-} \to c l_{2}$
$∴ m o l e s o f e^{-} = 0.5 \times 2 = 1$
$A l^{3 t} + 3 e^{-} \to A C$
$∴$ 1 mole of Ac requires 3 moles of e
$∴$ 1 mole $e^{-}$ reduces $\frac{1}{3}$ moles of AC
$m o l e o f A C = \frac{1}{3} \times 27 = 9$

Suggest Corrections

Similar questions

0.5 A

A l C l_{3}

(a t . w t . = 27)

0.5 A

A l C l_{3}

On passing 0.1 Faraday of electricity through aluminium chloride, the

amount of aluminium metal deposited on cathode is

[MP PMT 1991]

11.2

Join BYJU'S Learning Program