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10800 C of electricity through the elec...

`10800 C` of electricity through the electrolyte deposited `2.977 g` of metal with atomic mass `106.4 g mol^(-1)`. Find the charge on the metal cations.

Text Solution

Verified by Experts

The correct Answer is:
4
`M^(n+) + n e rarr M`
For `M, (w)/(E) = (it)/(96500)" " ( :' Q = i xx t)`
`(2.977)/(106.4//n) = (10800)/(96500)`
`:. n = (106.4 xx 10800)/(2.977 xx 96500) = 4`
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Knowledge Check

  • 10800 C of electricity passed through the electrolyte deposited 2.977g of metal with atomic mass 106.4 g mol^(-1) . The charge on the metal cation is

    A
    `+4`
    B
    `+3`
    C
    `+2`
    D
    `+1`

  • 10800 C of electricity passed through an electrolyte deposited 2.977 g of metal with atomic mass 106.4 " g mol"^(-1) The charge on the metal cation is :

    A
    `+ 4`
    B
    `+3`
    C
    `+2`
    D
    `+1`

  • On passingle C ampere of electricity through a electrolyte solution for t second, m gram metal deposits on cathode. The equivalent weight E of the metal is

    A
    `E=(Cxxt)/(mxx96500)`
    B
    `E=(Cxxm)/(txx96500)`
    C
    `E=(96500xxm)/(Cxxt)`
    D
    `E=(Cxxtxx96500)/(m)`

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