Question

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###Chemistry Quest!!!!!!

Question!!!!

=> Silver (atomic mass = 108 u) has a density of 10.5$g\times cm&{-3}$. The no of silver atoms on a surface of area $10^{-12}m^{2}$ can be expressed in scientific notation as $y\times 10^{x}$is...

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Answer 1

hello Bhai here is ur answer ==========

============================✌ ✌

d = mass/V

⇒ 10.5 g/cc means in 1 cc

⇒ 10.5 g of Ag is present.

Number of atoms of Ag in 1 cc

⇒ 10.5/108 × Na

In 1 cm, number of atoms of Ag = (10.5/108 × Na)^1/3

In 1 cm², number of atoms of Ag = (10.5/108 × Na)^2/3

In 10^ -12 m² or 10^ -8 cm², number of atoms of Ag = (10.5/108 × Na)^2/3 × 10^ -8

⇒ ((1.05 × 6.022 × 10^24)/108)^2/3 × 10^ -8

⇒ 1.5 × 10^7

So, x = 7...............

✌ ✌.......

Answer 2

1) We assume that unit cell is a cube.

Hence, approximate volume of the atom is:

V = (408.6 pm)³ = 68217388.06 pm³ = (68217388.06 pm³)(10^-30 cm³/1pm³) = 6.821 x 10^-23 cm³

As we know that density of the silver is 10.5 g/cm³ then

mass of Ag atom = density *Volume = (10.5 g/cm³)(6.821x10^-23 cm³) = 7.162 x 10^-22 g

2) As we know that atomic mass of silver is equal to 107.87 uma and that mass of a single atom is 7.162 x 10^-22 g hence,

Estimated Avogadro number is:

107.87 u / 7.162 x 10^-22 g/u = 1.506 x 10^23 átoms per mole of Silver

Hope it helps!