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Silver (atomic weight 108 g mol^(-1)) ha...

Silver (atomic weight `108 g mol^(-1))` has a density of `10.5 g cm^(-3)`. The number of silver atoms on a surfaces of area `10^(-12) m^(2)` can be expressed in scientific notation as `Y xx 10^(-x)`, The value of `x` is …….

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The correct Answer is:
7
`d=("mass")/(V) rArr 10.5 g//cc" means in 1 cc"rArr 10.5g` of Ag is present
Number of atoms of Ag in 1 cc `rArr (10.5)/(108) xx N_(A)`
In 1 cm, number of atoms of Ag `=sqrt((10.5)/(108) N_(A))`
In `1 cm^(2)`, number of atoms of Ag `=((10.5)/(108) N_(A))^(2//3)`
In `10^(-12)m^(2) or 10^(-8) cm^(2),` number of atoms of
`Ag=((10.5)/(108) N_(A))^(2//3) xx 10^(-8) =((1.05 xx 6.022 xx 10^(24))/(108))^(2//3) xx 10^(-8)`
Hence x=7 `=1.5 xx 10^(7)`
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Knowledge Check

  • Weight of 6.022 xx 10^(20) atoms of silver (at. mass 108 u) is

    A
    `108 xx 10^(3)g `
    B
    108 g
    C
    0.108 g
    D
    10.8 g

  • The number of silver atoms present in a 90% pure silver wire weighing 10 g is (Ag = 108)

    A
    `5.57xx10^(22)`
    B
    `0.62xx10^(23)`
    C
    `5.0 xx10^(22)`
    D
    `6.2 xx10^(29)`

  • How much time is required to deposit 1 xx 10^(-3) cm thick layer of silver (density is 1.05 g cm^(-3) ) on a surface of area 100 cm^(2) by passing a current of 5 A through AgNO_(3) solution ?

    A
    125 s
    B
    115 s
    C
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