You visited us 1 times! Enjoying our articles? Unlock Full Access!

Faraday's First Law

The charge re...

Question

The charge require to deposit $9 g$ of $A l$ from $A l^{3 +}$ solution is: (At. wt. of $A l = 27.0$ )

3216.3 C

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

96500 C

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

9650 C

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

32163 C

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

Open in App

Solution

The correct option is B $96500 C$
Half Cell reaction:-
$A l^{+ 3} + 3 e^{-} \to A l_{(s)}$ .

(i) Mole ratio $= \frac{m o l e s o f A l f o r m e d}{m o l e s o f e l e c t r o n s} = \frac{1}{3}$ .
(ii) Moles of $A l$ formed $= \frac{C h a r g e (Q)}{96500 C / m o l e^{-}} \times m o l e r a t i o$ .
$∴ \frac{W}{M M} = \frac{Q (C)}{96500 C / m o l e^{-}} \times \frac{1}{3}$ [ $m o l e s = \frac{W}{M M} = \frac{W e i g h t}{m o l a r m a s s}$ ].
$∴ \frac{9}{27} = \frac{Q (C)}{96500 C / m o l e^{-}} \times \frac{1}{3}$
$∴ Q (C) = \frac{9 \times 96500 \times 3}{27} = 96500 C$ .

Suggest Corrections

Similar questions

9 g

A l

A l^{3 +}

40.5 g

27.0 g / m o l e

A l_{2}

S O_{4})_{3}

36 g

A l

A l {(N O_{3})}_{3}

[A t . w t . o f A l = 27]

90 g

A l

A l^{3 +} + 3 e \to A l

1

= 27

A l C l_{3}

Join BYJU'S Learning Program