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13.5 gmof aluminium when changes to Al^(...

13.5 gmof aluminium when changes to `Al^(+3)` ion in solution, will lose: `[Ai = 27, N_(A) = 6 xx 10^(23)]`

A

`18.0xx10^23` electrons

B

`6.023xx10^23` electrons

C

`3.01 xx 10^23` electrons

D

`9.1xx10^(23)` electrons

Text Solution

Verified by Experts

The correct Answer is:
D
13.5 g of Al=13.5/27=0.5 mol of Al
`Al to Al^(3+)+3e^-`
3 moles of electrons are lost by 1 mole of Al.
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Knowledge Check

  • Molar mass of electron is nearly : (N_(A) = 6 xx 10^(23))

    A
    `9.1 xx 10^(-31)kg mol_(-1)`
    B
    `9.1 xx 10^(-31)gm mol^(-1)`
    C
    `54.6 xx 10^(-8)`gm `"mol"^(-1)`
    D
    `54.6 xx 10^(-8)kg "mol"^(-1)`

  • 6.02 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is (Avogadro constant, N_(A) = 6.02 xx 10^(23)mol^(-1))

    A
    `0.001 M`
    B
    `0.1M`
    C
    `0.002M`
    D
    `0.01`

  • 1.19gm mixture of Cu, Zn and Al contains 50u% copper by weight and 3 xx 10^(21) atoms of zinc. Find gram atoms of aluminum in given mixture? (At. Mass of Al= 27, Cu=63.5, Zn=65.0) N_(A) = 6 xx 10^(23)

    A
    0.001
    B
    0.27
    C
    0.01
    D
    0.027
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