Question

Q.33 Calculate number of atoms in the following
(1) 1g of Au(s) (II) 1g of Na (s) (III) 1g of Ll(s)
(Given Atomic mass of Au = 100 u, Na=23 u, Li=7 u)​

Answer 1

Given :

• Atomic mass of Au = 100
• Atomic mass of Na = 23
• Atomic mass of Li = 7
• 1g of Au
• 1g of Na
• 1g of Li

To find :

• Number of atoms in given substances

Solution :

(I) 1g of Au

Number of moles is calculated by ,

$\boxed {\rm{n = \frac{given \: weight}{atomic \: weight} }}$

⇒ Number of moles of Au = 1/100

⇒ Number of moles of Au = 0.01

Number of atoms in the given substance is given by ,

$\boxed {\rm{number \: of \: atoms = n \times N_A \: }}$

Here ,

• n is number of moles
• $\sf{N_A}$ is avogadro number (=6.023 × 10²³ )

⇒ Number of atoms in given Au = 0.01 × 6.023 × 10²³

⇒ Number of atoms in given Au = 10⁻² × 6.023 × 10²³

⇒ Number of atoms in given Au = 6.023 × 10²¹

Hence , The number of atoms in given Au is 6.023 × 10²¹

$\rule{300}{1}$

(II) 1g of Na

Number of moles of Na = 1/23

⇒ Number of moles of Na = 0.04

Now , Number of atoms is given by

Number of Atoms in given Na = 0.04 × 6.023 × 10²³

⇒ Number of atoms in given Na = 4 × 10⁻² × 6.023 × 10²³

⇒ Number of atoms in given Na = 24.092 × 10²¹

Hence , The number of atoms in given Na is 24.092 × 10²¹

$\rule{300}{1}$

(III) 1g of Li

Number of moles of Na = 1/7

⇒ Number of moles of Na = 0.14

Now , Number of atoms is given by

Number of Atoms in given Na = 0.14 × 6.023 × 10²³

⇒ Number of atoms in given Na = 14 × 10⁻² × 6.023 × 10²³

⇒ Number of atoms in given Na = 84.322 × 10²¹

Hence , The number of atoms in given Li is 84.322 × 10²¹

$\rule{300}{1}$

Answer 2

Answer:

Given :

Atomic mass of Au = 100

Atomic mass of Na = 23

Atomic mass of Li = 7

1g of Au

1g of Na

1g of Li

To find :

Number of atoms in given substances

Solution :

(I) 1g of Au

Number of moles is calculated by ,

\boxed {\rm{n = \frac{given \: weight}{atomic \: weight} }}

n=

atomicweight

givenweight

⇒ Number of moles of Au = 1/100

⇒ Number of moles of Au = 0.01

Number of atoms in the given substance is given by ,

\boxed {\rm{number \: of \: atoms = n \times N_A \: }}

numberofatoms=n×N

A

Here ,

n is number of moles

\sf{N_A}N

A

is avogadro number (=6.023 × 10²³ )

⇒ Number of atoms in given Au = 0.01 × 6.023 × 10²³

⇒ Number of atoms in given Au = 10⁻² × 6.023 × 10²³

⇒ Number of atoms in given Au = 6.023 × 10²¹

Hence , The number of atoms in given Au is 6.023 × 10²¹

\rule{300}{1}

(II) 1g of Na

Number of moles of Na = 1/23

⇒ Number of moles of Na = 0.04

Now , Number of atoms is given by

Number of Atoms in given Na = 0.04 × 6.023 × 10²³

⇒ Number of atoms in given Na = 4 × 10⁻² × 6.023 × 10²³

⇒ Number of atoms in given Na = 24.092 × 10²¹

Hence , The number of atoms in given Na is 24.092 × 10²¹

\rule{300}{1}

(III) 1g of Li

Number of moles of Na = 1/7

⇒ Number of moles of Na = 0.14

Now , Number of atoms is given by

Number of Atoms in given Na = 0.14 × 6.023 × 10²³

⇒ Number of atoms in given Na = 14 × 10⁻² × 6.023 × 10²³

⇒ Number of atoms in given Na = 84.322 × 10²¹

Hence , The number of atoms in given Li is 84.322 × 10²¹

\rule{300}{1}