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BITSAT 2018 Paper

Section: Chemistry

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Question : 64 of 150

Marks: +1, -0

On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of

\mathrm{CuCl}_2

, all copper of the solution was deposited at cathode. The strength of

\mathrm{CuCl}_2

solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500

\mathrm{Cmol}^{-1}

)

0.01 N
0.01 M
0.02 M
0.2 N

Solution:

By Faraday's 1st Law,

\frac{W}{E}

=

\frac{q}{96500}

(where q = it = charge of ion)

we know that no of equivalent

=

\frac{W}{E}

=

\frac{it}{96500}

=

\frac{1 \times 965}{96500}

=

\frac{1}{100}

(where i = 1 A, t = 16×60+5 = 965 sec.)

Since, we know that

Normality =

\frac{\text{no. of equivalent}}{\text{Volume (in litre)}}

=

\frac{\frac{1}{100}}{1}

= 0.01 N

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