Ncert Exercises & Solutions

What is the average mass density of a sodium atom assuming its size to be about 2.5 angstrom. (Use the known values of Avogadro’s number and the atomic mass of sodium).
1. $4.67\times10^{3}~\text{kg/m}^3$
2. $4.67\times10^{3}~\text{g/cm}^3$
3. $4.67\times10^{10}~\text{kg/m}^3$
4. $4.67\times10^{10}~\text{g/cm}^3$

^{$H e r e,$

$r = \frac{d}{2} = \frac{2.5 \times 10^{- 10}}{2} = 1.25 \times 10^{- 10} m$

$A t o m i c v o l u m e = \frac{4}{3} π r^{3} \times N_{A} = \frac{4}{3} \times \frac{22}{7} \times {(1.25 \times 10^{- 10})}^{3} \times 6.023 \times 10^{23}$

$= 4.93 \times 10^{- 6} m^{3}$

$A v e r a g e m a s s d e n s i t y = \frac{m a s s}{v o l u m e} = \frac{23 \times 10^{- 3}}{4.93 \times 10^{- 6}}$

$= 4.67 \times 10^{3} k g m^{- 3} .$}

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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