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Byju's Answer
Standard XII
Chemistry
Faraday's First Law
How many gram...
Question
How many grams of water will be electrolysed by
96500
coulomb charge?
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Solution
96500
C
charge,
H
2
O
⟶
H
2
+
1
2
O
2
∴
mole of
e
−
transferred per mole of water
=
2
∴
Chagre per mole of water
=
2
×
1.6
×
10
−
19
C
×
6.022
×
10
23
=
19.27
×
10
4
∴
no. of moles of water
=
96500
19.27
×
10
4
=
0.50078
⟹
No. of grams of
H
2
O
=
0.50078
×
18
=
9.014
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Q.
If one Faraday was to be
48250
coulomb instead of
96500
coulomb, what will be charge of an electron?
Q.
If 1 Faraday were to be 60230 coulombs instead of 96500 coulombs, what will be the charge of an electron?
(1 Faraday= total charge on the collection of 1 mole of electrons)
Q.
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode.
[Atomic mass: Na = 23, Hg = 200; 1 Faraday = 96500 coulombs]
The total charge (coulombs) required for complete electrolysis is:
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How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane?
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=
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]
Q.
How many moles of electrons will together contribute to the charge of 289500 coulomb?
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