Question

An element has bcc structure with a cell edge of 300pm. the density of the element is 6.3 gcm-3. How many atoms are present ¡n 303 g of the element

Answer 1

Answer:

The expression for density (d) is given below:

d=

a

3

×N

A

n×M

For BCC,

n=2

a=290 pm =290×10

−10

cm

d=6.8 g cm

−3

6.8=

(290×10

−10

)

3

×6.023×10

23

2×M

M=

2

6.8×6.023×10

23

×24.4×10

−30

=50

∴ Number of atoms present =

50

6.023×10

23

×200

=2.4×10

24

atoms

Hence, the number of atoms present in 200 g of the element is 2.4×10

24

.

Explanation:

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Answer 2

Answer:

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