Formula mass of `NaCl` is `58.45g mol^(-1)` and density of its pure form is `2.167g cm^(-3)`. The average distance between adjacent sodium and chloride ions in the crystal is `2.814xx10^(-8)cm`. Calculate Avogadro constant.

Answer

`6.06 xx 10^(23)`

Explanation

Edge of unit cell = `2 xx 2.814 xx 10^(-8) cm`
`= 5.628 xx 10^(-8) cm`
`Z = 4`
If `N_A ` is Avogardro number,
Density = `(Z xx M)/(a^3 xx N_A)`
`M = 202 + 37 = 239`
`2.167 = (4 xx 58.5)/((5.628 xx 10^(-8))^(3) xx N_A)`
`N_A = (4 xx 58.5)/((5.628 xx 10^(-8))^(3) xx 2.167)`
`= 6.06 xx 10^(23)` .