Question

Sodium chloride crystallises in face centred cubic structure. Its density is 2.165 g/cm^3. If the distance between Na+ and its nearest Cl- is 281 pm. Find its Avogardo Number .no spam

Answer 1

Answer : 6.089 ×$10^{23}$

Given,

Density of NaCL = 2.165 g/cm^3

radius of Na = radius of Cl(a') = 281 pm

1 pm = 10^-10cm

281 pm = 281 * 10^-10 cm

As we know,

diameter = 2 radius  

a , edge of the unit cell (diameter) = 2 * a'

= 2* 281 * 10^-12 m  

= 562 *  10^-12 m  

As we know the formula,

density (raw)  = $\frac{z*m}{a^{3}*N_{a}}$

here ,

z ,  number of atoms in FCC  = 4  

m = 58.50

m = molar mass of NaCl = 23 + 35.5 = 58.50

$N_{a}$ = avogardo number

On putting the values in the formula,

2.165 = $\frac{4*58.50}{(562*10^{-10})^{3}*N_{a}}$

$N_{a}$ = 6.089 × $10^{23}$

                                   (nearer to 6.023 ×$10^{23}$)

                           

Answer 2

Density of NaCL = 2.165 g/cm^3

radius of Na = radius of Cl(a') = 281 pm

1 pm = 10^-10cm

281 pm = 281 * 10^-10 cm

As we know,

diameter = 2 radius

a , edge of the unit cell (diameter) = 2 * a'

= 2* 281 * 10^-12 m

= 562 * 10^-12 m

na= 10.089× 10^23