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Question

Chemistry Question on Conductance

When same quantity of electricity is passed through aqueous AgNO3AgNO_3 and H2SO4H_2SO_4 solutions connected in series, 5.04×102g5.04 \times 10^{-2} \, g of H2H_2 is liberated. What is the mass of silver (in grams) deposited? (E wts. of hydrogen = 1.008, 1.008, silver = 108108)

A

54

B

0.54

C

5.4

D

10.8

Answer

5.4

Explanation

Solution

Given, weight of hydrogen liberated
=5.04×102g=5.04 \times 10^{-2} g
E wt. of hydrogen =1.008=1.008
E wt. of silver =108=108
Weight of silver deposited, w=?w =?
According to Faraday's second law of electrolysis
 weight of silver deposited  weight of hydrogen liberated \frac{\text { weight of silver deposited }}{\text { weight of hydrogen liberated }}
= e wt .of silver  e wt. of hydrogeon =\frac{\text { e wt .of silver }}{\text { e wt. of hydrogeon }}
w5.04×102=1081.008\frac{w}{5.04 \times 10^{-2}}=\frac{108}{1.008}
W=108×5.04×1021.008W=\frac{108 \times 5.04 \times 10^{-2}}{1.008}
=5.4g=5.4\, g