By passing .1 faraday of electricity through fused nacl. The amount of chlorine liberated is
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Answered by
8
On passing 0.1F of electricity in NaCl,
the amt of chloride liberated = 35.45*0.1 =3.545
hence it is option C
Answered by
6
Answer:
Explanation:W of cl2 =no.of faradays (into).35.5
=0.1(into).35.5gm
=3.545grams
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