On passing 0.01 Faraday of elctricity through a solution of aluminum chloride, the weight of Aluminium metal $(a t . w t . = 27)$ deposited on cathode is:

0.9 g m

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

0.09 g m

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

2.7 g m

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

0.27 g m

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

Open in App

Solution

The correct option is B $0.09 g m$
${A l}^{3 +} + 2 e^{-} \to A l$
Mole of electricity $= 0.01 F$
$3 m o l \to 27 g o f A l$
$0.01 F \to \frac{1}{3} \times 27 \times 0.01$
$= 0.09 g m$

Suggest Corrections

Similar questions

On passing 0.1 Faraday of electricity through aluminium chloride, the

amount of aluminium metal deposited on cathode is

[MP PMT 1991]

965

amp current is passed through molten metal chloride for one minute and

40

seconds during electrolysis. The mass of metal deposited is

9 g m

at the cathode. The valency of metal atom (at.wt

= 27

) is:

Three Faradays of electricity was passed through an aqueous solution of magnesium bromide. The weight of magnesium metal deposited at the cathode in grams is:

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:

3

Faradays of electricity was passed through an aqueous solution of iron

(I I)

bromide. The weight (in g) of iron metal (atomic weight

= 56

amu) deposited at the cathode is:

Join BYJU'S Learning Program

On passing 0.01 Faraday of elctricity through a solution of aluminum chloride, the weight of Aluminium metal (at.wt.=27) deposited on cathode is:

The correct option is B 0.09gmAl3++2e−→AlMole of electricity =0.01F3mol→27gofAl0.01F→13×27×0.01 =0.09gm

On passing 0.01 Faraday of elctricity through a solution of aluminum chloride, the weight of Aluminium metal $(a t . w t . = 27)$ deposited on cathode is:

The correct option is B $0.09 g m$
${A l}^{3 +} + 2 e^{-} \to A l$
Mole of electricity $= 0.01 F$
$3 m o l \to 27 g o f A l$
$0.01 F \to \frac{1}{3} \times 27 \times 0.01$
$= 0.09 g m$