Question

please fast answer guys

Answer 1

Hi there,

Here's the answer:

°•°•°•°<><><>==<><><>°•°•°•°•°•°

First,

Points to Remember:

→ Density  $D=$$\frac{z*M}{N_{A} *a^{3} }$

=> $M=\frac{D*N_{A} *a^{3} }{z}$

→ For BCC Lattice, z = 2

Here,

Z= No. of atoms present in the unit cell.

M= Molar Mass

a= edge length

$N_{A}$= Avogardo constant= 6.022*10^(23).

°•°•°•°<><><>==<><><>°•°•°•°•°•°

Given,

a= 250 pm= 250*$10^{-12}$m

=> a= 250*$10^{-10}$cm

D= 7.2 g/cm³

Substitute values in M

∵ z=2 for BCC

$a^{3} = (250*10^{-12} )^{3}$

$=15625*10^{3}*10^{-30}$

$M=\frac{D*N_{A} *a^{3} }{2}$

$M = \frac{7.2*6.023*10^{23}*15625*10^{3}*10^{-30}}{2}$

∴ Molar mass = 33 grams

°•°•°•°<><><>==<><><>°•°•°•°•°•°

;)

Hope it helps

Answer 2

Hi there,



Here's the answer:



°•°•°•°<><><>==<><><>°•°•°•°•°•°



First,



Points to Remember:



→ Density  D=D= \frac{z*M}{N_{A} *a^{3} }NA​∗a3z∗M​



=> M=\frac{D*N_{A} *a^{3} }{z}M=zD∗NA​∗a3​



→ For BCC Lattice, z = 2



Here,



Z= No. of atoms present in the unit cell.



M= Molar Mass



a= edge length



N_{A}NA​ = Avogardo constant= 6.022*10^(23).



°•°•°•°<><><>==<><><>°•°•°•°•°•°



Given,



a= 250 pm= 250*10^{-12}10−12 m

=> a= 250*10^{-10}10−10 cm

D= 7.2 g/cm³


Substitute values in M

∵ z=2 for BCC

a^{3} = (250*10^{-12} )^{3}a3=(250∗10−12)3

=15625*10^{3}*10^{-30}=15625∗103∗10−30


M=\frac{D*N_{A} *a^{3} }{2}M=2D∗NA​∗a3​

M = \frac{7.2*6.023*10^{23}*15625*10^{3}*10^{-30}}{2}M=27.2∗6.023∗1023∗15625∗103∗10−30​



∴ Molar mass = 33 grams



°•°•°•°<><><>==<><><>°•°•°•°•°•°



;)



Hope it helps