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Grade: 12
        
Same electric charge is passed through aqua solution of HCL and CuSo4. if 12 gram of H2 is liberated find the mass of copper deposited
1.    402 g
2.    360 g
3.    381 g 
4.    369 g
3 months ago

Answers : (2)

Arun
22825 Points
							
W1 = (A1 X I X t) ÷ (n1 x 96500)
W1 is mass of H2
12 = (1 X I X t) ÷ (2 x 96500)
12 x 193000 = I X t ---------------- equation 1
W2 = (A X I X t) ÷ (n2 x 96500)
W2 is the mass of Cu
W2 = (63.5 X 12 x 193000) ÷ 193000 = 762g/2 = 381
 
3 months ago
Vikas TU
9225 Points
							
2 mole of electron will liberate 1 mole of H2 gas 
Mole of H2 liberated = 12/2 = 6 
To liberate 6 mole of e we require 12 mole of e 
To liberate 2 mole of e we need 1 mole of cu 
12 mole of e will deposit 6 mole of cu 
Mass of cu = 6* 63.5 = 381 
3 months ago
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