$0.5 f a r a d a y$ of electricity was required to deposit all the copper in $500 m L$ of a copper sulphate solution. What is the normality of the copper sulphate solution?

0.5 N

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1 N

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1.5 N

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2 N

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Solution

The correct option is B 1 N
By Faraday's law , $ω$ =Zq
q= charge in faraday
no. of equivalents = $\frac{ω}{Z}$ =0.5
0.5 = Normality $\times$ Volume (in L)
0.5 =Normality $\times$ 500 $\times 10^{- 3}$
Therefore Normality = 1N.

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0.5 faraday of electricity was required to deposit all the copper in 500 mL of a copper sulphate solution. What is the normality of the copper sulphate solution?

The correct option is B 1 NBy Faraday's law , ω =Zqq= charge in faraday no. of equivalents = ωZ =0.50.5 = Normality × Volume (in L)0.5 =Normality × 500 ×10−3Therefore Normality = 1N.

$0.5 f a r a d a y$ of electricity was required to deposit all the copper in $500 m L$ of a copper sulphate solution. What is the normality of the copper sulphate solution?

The correct option is B 1 N
By Faraday's law , $ω$ =Zq
q= charge in faraday
no. of equivalents = $\frac{ω}{Z}$ =0.5
0.5 = Normality $\times$ Volume (in L)
0.5 =Normality $\times$ 500 $\times 10^{- 3}$
Therefore Normality = 1N.