Ncert Exercises & Solutions

Estimate the average mass density of a sodium atom assuming its size to be about $2.5 \overset{0}{A}$ (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: $970 k g m^{- 3} .$ Are the two densities of the same order of magnitude? If so, why?

^{$H e r e,$
$r = \frac{d}{2} = \frac{2.5 \times 10^{- 10}}{2} = 1.25 \times 10^{- 10} m$
$A t o m i c v o l u m e = \frac{4}{3} π r^{3} \times N_{A} = \frac{4}{3} \times \frac{22}{7} \times {(1.25 \times 10^{- 10})}^{3} \times 6.023 \times 10^{23}$
$= 4.93 \times 10^{- 6} m^{3}$
$A v e r a g e m a s s d e n s i t y = \frac{m a s s}{v o l u m e} = \frac{23 \times 10^{- 3}}{4.93 \times 10^{- 6}}$
$= 4.67 \times 10^{3} k g m^{- 3} .$}

The density of sodium in the crystalline phase is given

970 k g m^{- 3} .

Hence, the density of the sodium atom and the density of sodium in its crystalline phase is not in the same order. This is because, in the solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.

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