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Estimate the averaage atomic mass density of a sodium atom, assuming its size ot be 2.5 Å. Compare it with density of sodium in its crystalline phase (970 kg m^(-3)). Are the two denities of the same order of magnitude ? If so, why ?

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Answer Text

Solution :
It is given that radius of sodium atom, `R=2.5 A=2.5xx10^(-10) m` <br> Volume of one mole atom of sodium, `V=NA. 4//3 pi R^(3)` <br> `V=6.023xx10^(23)xx-4//3xx3.14xx(2.5xx10^(-10))^(3) m^(3)` and mass of one mole atom of sodium, `M=23 g=23xx10^(-3) kg` <br> `:.` Average mass density of sodium atom, `p=M//V` <br> `=(23xx10^(-3)//6.023xx10^(23)xx4//3xx3.14xx(2.5xx10^(-10)))` <br> `=6.96xx10^(2) kg m^(-3)=0.7xx10^(-3) kg m^(-3)` <br> the density of sodium in its crystalline phase `=970 kg m^(-3)` <br> `=0.97xx10^(3) kg m^(-3)` <br> Obviously the two densities are of the same order of magnitude `(=10^(3) kg m^(-3))`. It is on account of the fact that in solid phase atoms are tightly packed and so the atomic mass density is close to the mass density of solid.

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