Question

A current of 2 A passed for 1 hour through a solution of molten aluminium chloride has a efficiency of 50%.Find the amount of Al deposit.
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Answer 1

Answer:

daksh ki mummy ek number ki jhoothi usko peeto

Answer 2

The amount of aluminum deposit is 0.335g.

Given:

A current of 2 A passed for 1 hour through a solution of molten aluminum chloride has an efficiency of 50%.

To Find:

The amount of Al deposit.

Solution:

To find the Al deposit we will follow the following steps:

$AlCl₃→ {Al}^{3 + } + 3{Cl}^{ - }$

Here, the electrons exchanged are 3.

Also,

$w = \frac{m \: \times it}{nf}$

Here,

w is the amount of Al deposits on passing the current of 2A.

i = 2A

t = 1 hour = 60 × 60 = 3600-sec

(1 h = 60min and 1 min = 60 sec)

n = 3

f = 96500C

m = molar mass of aluminum = 27g

Now,

In putting values we get,

$w = \frac{27 \times 2 \times 3600}{3 \times 96500} = 0.67g$

Efficiency is 50%

Also,

$efficiency = \frac{aluminium \: actually \: deposit}{aluminium \: deposit \: on \: complete \: utilisation \: of \:current \: } \times 100$

$Amount \: of \: aluminum \: deposit = \frac{efficiency \times aluminium \: deposit \: on \: complete \: utilisation \: of \:current \:}{100}$

$= \frac{0.67 \times 50}{100} = \frac{0.67}{2} = 0.335g$

Henceforth, the amount of aluminum deposit is 0.335g.

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