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Physics
The value of Faraday constant is
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Q. The value of Faraday constant is
Chhattisgarh PMT
Chhattisgarh PMT 2005
A
$9.65 \times 10^{4} C$
B
$9.65 \times 10^{4} J$
C
$6.03 \times 10^{27} J$
D
$1.6 \times 10^{-19} C$
Solution:
Faraday constant $F=N e$
$F=6.023 \times 10^{23} \times 1.6 \times 10^{-19} C=96368\, C $
$=9.65 \times 10^{4}\, C$