Question

A current of 0.5 amperes is passed for half an hour through a voltmeter having copper sulphate solution. The weight of copper deposited is: Option 1: 0.350 gm Option 2: 1.20 gm Option 3: none of these Option 4: 0.297 gm​

Answer 1

Explanation:

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Answer 2

Given: Current passed, I = 0.5 A

           Time taken, t = 1/2 hr

To Find: the weight of copper deposited, m.

Solution:

To calculate m, the formula used:

• Charge = current x time
• Q = I x t                                                      ⇒ 1
• Also, charge = number of electrons flow x number of moles x Faraday's constant
• Q = n xN x F
• Number of moles = weight deposited / molar mass
• N = m / M
• Q = n x m x F / M                                        ⇒2

Applying the formula 1:

Q = 0.5 x (1/2)

Convert 1/2 hr into seconds;

1 hr = (60 x 60 ) s

0.5 hr = 0.5x 3600

          = 1800 s

∴ Q = 0.5 x 1800

      = 900 C                              ⇒3

Now applying the formula 2:

Q = n x m x F / M

here,n = 2

       M = 63.5 g

       F = 96500 C

Q   = 2 x m x 96500 / 63.5

      = m x 193040 / 63.5

      = m x 3040                          ⇒ 4

Equating equations 3 and 4:

900 = m x 3040

m = 900 / 3040

   =  90 / 304

   = 0.297

m = 0.297 gm

Hence, the weight of copper deposited is 0.297 gm ( option-4).