Question

.The density of mercury is 13.6 g/mL. Calculate the diameter of an atom of mercury
assuming that each atom of mercury is occupying a cube of edge-length equal to the
diameter of the mercury atom. (Hg = 200)
​

Answer 1

Answer:

Avogadro's number (NA​)=6.023×1023

Atomic weight of Hg=200

∵ In 1 gm of Hg number of atoms present

=2×1026.023×1023​

=3.012×1021

∵ Densiry of mexcury =13.6g∣c.c

∴ mass of 3.012×1021 atoms =3.012×10211​

volumes of 1 atoms of Hg

3.012×1021×13.6g∣c.c1​=3012×1021×136103×10​

⇒3012×136cc10−17​=409632cc10−17​

=2.44×1023c.c

Since each mercury atoms occuples a cube of edge length equals to its diameter.

Diameter of one Hg atom =(2.44×10−23)1/3cm

=(2.44×10−24)1/3cm

=2.905×10−8cm

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Avogadro's number (N

A

)=6.023×10

23

Atomic weight of Hg=200

∵ In 1 gm of Hg number of atoms present

=

2×10

2

6.023×10

23

=3.012×10

21

∵ Densiry of mexcury =13.6g∣c.c

∴ mass of 3.012×10

21

atoms =

3.012×10

21

1

volumes of 1 atoms of Hg

3.012×10

21

×13.6g∣c.c

1

=

3012×10

21

×136

10

3

×10

⇒

3012×136cc

10

−17

=

409632cc

10

−17

=2.44×10

23

c.c

Since each mercury atoms occuples a cube of edge length equals to its diameter.

Diameter of one Hg atom =(2.44×10

−23

)

1/3

cm

=(2.44×10

−24

)

1/3

cm

=2.905×10

−8

cm