Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol-1]
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Correct Answer: Option D
Explanation:
m = \(\frac{Mm \times Q}{96500n
}\)
where Q = IT
M = Mm × \(\frac{Q}{96500n}\)
where m = mass
Mm = Molar mass
Q = Quantity of electricity
n = number of change= +1
\(\frac{M}{Mm}\) = mole = \(\frac{mass}{Molarmass}\)
\(\frac{M}{Mm}\) = \(\frac{Q}{96500n}\)
= \(\frac{9650}{96500n}\) × 1
= \(\frac{1}{10}\) = 0.1mol
m = \(\frac{Mm \times Q}{96500n
}\)
where Q = IT
M = Mm × \(\frac{Q}{96500n}\)
where m = mass
Mm = Molar mass
Q = Quantity of electricity
n = number of change= +1
\(\frac{M}{Mm}\) = mole = \(\frac{mass}{Molarmass}\)
\(\frac{M}{Mm}\) = \(\frac{Q}{96500n}\)
= \(\frac{9650}{96500n}\) × 1
= \(\frac{1}{10}\) = 0.1mol