menu search
brightness_auto
more_vert

An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element?

thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike

1 Answer

more_vert
 
verified
Best answer

Given: Types of unit cell = bcc, for bcc unit cell n = 2

Edge length (a) = 288 pm = 2.88 x 10-8 cm3

Mass of element (x) =200g

Density of element (r) = 7.2 gm/cm3

Number of unit cell in x g of element = \(\frac{x}{ρa^3}\)

= \(\frac{200}{7.2×(2.88×10^{-8})^3}\) = 1.16 x 1024

Number of atoms in x g of element = \(\frac{nx}{ρa^3}\)

= \(\frac{2x}{ρa^3}\)= 2 × 1.16 x 1024 = 2.32 x 1024

Ans. Number of unit cell in x g of element = 1.16 x 1024

Number of atoms in x g of element = 2.32 x 1024

thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike

Related questions

thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer

4.8k questions

4.3k answers

67 comments

387 users

...