Question

solve this .......☺️

Answer 1

HELLO DEAR,

we know:-Avogadro’s number = 6.023 * 10²³
Atomic mass of mercury(Hg) = 200

∵ In 1 g of Hg, the total number of atom

⇒ 6.023 *10²³/200

⇒ 6.023 *10²³/2 *10²

⇒ 3.0115 *10²¹

⇒ 3.012 * 10²¹

∵given:- Density of Mercury (Hg) = 13.6 g/c.c.

NOW,
mass of 3.012 * 10²¹ atoms ⇒ 1/3.012 *10²¹
and volume of 1 atom of mercury (Hg)

⇒ 1/(3.012 *10²¹ *13.6) c.c.

⇒ 10³*10/(3012 *10²¹* 136) c.c.

⇒ 10⁻¹⁷/(3012 *136) c.c.

⇒ 10⁻¹⁷/(409632) c.c.

⇒ 10⁵*10⁻²³/409632 c.c.

⇒2.44 *10⁻²³ c.c.

⇒2.44 *10⁻²³ c.c.

each mercury atom occupies a cube of edge length equal to its diameter, therefore,
Diameter of one Hg atom

⇒ (2.44 * 10⁻²³)^⅓ cm

⇒ (24.4 * 10⁻²⁴)^⅓ cm.

⇒ 2.905 * 10⁻⁸ cm

⇒ $\bold{\boxed{\sf{\large{2.91 }}}}$Å

I HOPE ITS HELP YOU DEAR,
THANKS