Calculate the mass of chlorine liberated by passing 28950 c of electricity through hcl solution
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On passing 0.1F of electricity in NaCl,
the amt of chloride liberated = 35.45*0.1 =3.545
hence it is option C
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ANSWER:-
On passing 0.1F of electricity in NaCl,
the amt of chloride liberated = 35.45*0.1 =3.545
hence it is option C
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