When X amperes of current is passed through molten $A l C l_{3}$ for 96.5 seconds. 0.09 grams of aluminium is deposited. What is the value of X?

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Solution

The correct option is A 10
Given $m = 0.9 g$

Oxidation number $= 3$

$5 = 96.5 s$

$i = X = ?$

$m = \frac{E}{F} i t$

$0.09 = \frac{(27) \times (965)}{3 \times 96500}$

$X = \frac{0.9 \times 3 \times 1000}{27} = 10 A$

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When X amperes of current is passed through molten AlCl3 for 96.5 seconds. 0.09 grams of aluminium is deposited. What is the value of X?

The correct option is A 10Given m=0.9gOxidation number =35=96.5 si=X=?m=EFit0.09=(27)×(965)3×96500X=0.9×3×100027=10A

When X amperes of current is passed through molten $A l C l_{3}$ for 96.5 seconds. 0.09 grams of aluminium is deposited. What is the value of X?

The correct option is A 10
Given $m = 0.9 g$

Oxidation number $= 3$

$5 = 96.5 s$

$i = X = ?$

$m = \frac{E}{F} i t$

$0.09 = \frac{(27) \times (965)}{3 \times 96500}$

$X = \frac{0.9 \times 3 \times 1000}{27} = 10 A$