Question

Question 2.27 Estimate the average mass density of a sodium atom assuming its size to be about 2.5. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m–3. Are the two densities of the same order of magnitude? If so, why?

Chapter Units And Measurements Page 37

Answer 1

Size of sodium atom= 2.5 A° = diameter of sodium atom
radius of sodium atom = 2.5A°/2
= 1.25 A°
Volume of sodium atom = 4/3 πr³
= 4/3 × 22/7 × (1.25A°)³
= 8.177 × 10^-30 m³ [ 1A° = 10^-10m]

Mass of sodium atom = mass of one mole of sodium atom/Avogadro constant
= 23 × 10-³/6.023 × 10²³ Kg
= 3.818 × 10^-26 Kg

Now,
Density = 3.818× 10^-26/8.177 × 10^-30
= 4.67 × 10³ Kg/m³
Density of sodium atom in solid state = 4.67 × 10³ Kg/m³ but in crystalline state density of sodium atom = 970 kg/m³ . hence, both are in different order. Becoz in solid phase atoms pack closely but in crystalline phase arrange a sequence , contain void so, dneisty in solid phase greater then crystalline phase .

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.