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The density of mercury is 13-6 g/cc. Cal...

The density of mercury is 13-6 g/cc. Calculate approximately the diameter of an atom of mercury, assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom.

Text Solution

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Suppose the length of the side of the cube is X cm, i.e., the diameter of one Hg atom. `therefore` volume occupied by 1 Hg atom = `X^3` cc
and mass of one Hg atom = `13.6 xx X^(3)` g.
Mass of one Hg atom `=("at. Wt")/("Av. Const") = 200/(6.022 xx 10^(23))g`
mass of 1 mole of atoms is the atomic weight in g, and 1 mole contains the Av. const, of atoms)
Hence, `13.6 xx X^(3) = 200/(6.022 xx 10^(23))`
`X^(3) = 200/(13.6 xx 6.022 xx 10^(23)) = 2.44 xx 10^(-23)`
`X =2.9 xx 10^(-8)` cm
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Knowledge Check

  • Water rises to height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6 g cm^(-3) and the angle of contact of mercury and water are 135^(@) and 0^(@) respectively, the ratio of surface tension of water and mercury is

    A
    `1:0:15`
    B
    `1:3`
    C
    `1:6.5`
    D
    `1.5:1`

  • A vessel contains oil (dersity 0.8 g/cc) over mercury (density 13.6 g/cc). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere is g/ccis:

    A
    `3.3 `
    B
    `6.4 `
    C
    `7.2 `
    D
    `12.8`

  • Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.5 cm in the same capillary tube. If the density of mercury is 13.6 gm//cc and its angle of contact is 135^(@) and density of water is 1 gm//cc and its angle of contact is 0^(@) , then the ratio of surface tensions of the two liquids is (cos 135^(@)=0.7 )

    A
    `1:14`
    B
    `5:34`
    C
    `1:5`
    D
    5:27`

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