Question

2.27 Estimate the average mass density of a sodium atom assuming its size to be about
2.5 A. (Use the known values of Avogadro's number and the atomic mass of sodium).
Compare it with the mass density of sodium in its crystalline phase : 970 kg m Are
the two densities of the same order of magnitude ? If so, why?​

Answer 1

Answer:

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Answer 2

$\huge\underline{Explanation•}$

$volume \: of \: Na \: atom \: = \: v \: = \: \frac{4}{3} \pi {r}^{3} \\ \: = \frac{4}{3} \times 3.14 \times ({2.5 \times 10 ^{ - 10} })^{3} \\ = 65.42 \times {10}^{ - 30} m ^{3}$

Now,

$volume \: of \: one \: mole \: of \: Na \: atom \: \: v = na \times v \\ = 6.023 \times {10}^{23} \times 65.42 \times {10}^{ - 30} \: {m}^{3} \\ = 3.95 \times {10}^{ - 5} \: {m}^{3}$

Mass of one mole of Na atom = 23 gm

$density \: of \: Na \: atoms \: = \frac{23 \times {10}^{ - 3} }{3.95 \times {10}^{ - 5} } = 584 \: kg \: {m}^{ - 3}$

Here , 584 and 970 are of same order.

Note:- Density of sodium in crystalline phase little differ from density of atoms because in crystalline molecules are arranged in sequence and in atoms closely packed.