Question #c5811

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Question #c5811

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Answer 1 · 2016-08-21T14:01:54

Following the law of equivalent proportion we can say that

$\frac{Equivalent wt of Zn}{Equivalent wt of Ag} = \frac{Displacing mass of Zn}{Displaced mass of Ag}$ $"Equivalent wt of Zn"/ "Equivalent wt of Ag"="Displacing mass of Zn"/"Displaced mass of Ag"$

$\Rightarrow \frac{E_{Zn}}{E_{Ag}} = \frac{1.04}{1.728}$ $=>E_"Zn"/E_"Ag"=1.04/1.728$

$\Rightarrow \frac{E_{Zn}}{108} = \frac{1.04}{1.728}$ $=>E_"Zn"/108=1.04/1.728$

$\Rightarrow E_{Zn} = \frac{1.04}{1.728} \times 108 = 65$ $=>E_"Zn"=1.04/1.728xx108=65$

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