Question

✯ Chemistry Question:-

An element has cube type unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of the unit cell is 24 × 10-²⁴ cm³ and density of the element 7.2 gram/cm³. Calculate the atoms present in 100 gram of the element.​

Answer 1

Answer:-

$\pink{\bigstar}$ The atoms present in 100 gram of the element is $\large\leadsto\boxed{\tt\purple{1.736 \times 10^{24} \: atoms}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Given:-

• Volume of unit cell = 24 × 10-²⁴ cm³

• Density of the element = 7.2 g/cm³

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• To Find:-

• The atoms present in 100 gram of the element.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Solution:-

We know,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{d = \dfrac{z \times M}{N \times a^3}}}}$

where,

• d = Density of the element

• z = Number of atoms in unit cell [z = 1 + 2 = 3]

• M = Atomic mass

• N = Avagadro Number [6.02 × 10²³]

• a³ = Volume of the unit cell

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Substituting in the Formula:-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf 7.2 = \dfrac{3 \times M}{6.02 \times 10^{23} \times 24 \times 10^{-24}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf M = \dfrac{7.2 \times 6.02 \times 10^{23} \times 24 \times 10^{-24}}{3}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf M = \dfrac{1040.256 \times 10^{-1}}{3}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf M = \dfrac{104.0256}{3}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ $\large{\bf\red{M = 34.69}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Now,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➠ 34.69 gram of element contains atoms = 6.02 × 10²³

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➠ 100 gram of element contains atoms =

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf \dfrac{6.02 \times 10^{23} \times 100}{34.69}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf \dfrac{6.02 \times 10^{25}}{34.69}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf 0.17353 \times 10^{25}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf 1.7353 \times 10^{24}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ $\large{\bf\pink{1.736 \times 10^{24} \: atoms}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the atoms present in 100 gram of the element is 1.736 × 10²⁴ atoms.

Answer 2

⠀⠀☆ Question Given :

• An element has cube type unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of the unit cell is 24 × 10-²⁴ cm³ and density of the element 7.2 gram/cm³. Calculate the atoms present in 100 gram of the element.

⠀⠀☆ Required Solution :

⠀⠀✫ Values Given to us :

• ➷ The volume of the unit cell is 24 × 10-²⁴ cm³

• ➷ Density of the element 7.2 gram/cm³.

⠀⠀✫ Formula Used Here :

• ➷ d = z × M / N × a³

☆ Putting values in Formula :

• ➷ 7.2 = 3 × M / 6.02 × 10 ²³ × 24 × 10 -²⁴

• ➷ M = 7.2 × 6.02 × 10²³ × 24 10-²⁴×

• ➷ M = 1040.256 × 10-¹

• ➷ M = 104.0256 / 3

⠀⠀☆ M = 34.69 ( Approx ) ☆

⚘ Finding for 34.69g Elements = 6.02 × 10²³

⠀✫ According to Question :

• ➷ 6.02 × 10²³ × 100 / 34.69

• ➷ 6.02 × 10^25 / 34.69

• ➷ 0.17353 × 10^25

⠀✫ Adjust the Expression :

• ➷ 1.7353 × 10²⁴

✫ Atom Present in 100g of element :

⠀⠀⠀➼ ★ 1.7353 × 10²⁴ ★

⠀⠀⠀⠀⠀⠀⠀(Answer)

___________________________

⠀⠀✯ Knowledge Booster ✯

• ❛ d ❜ ⇢ Density of the element

• ❛ z ❜ ⇢ Number of atoms in unit cell

• ❛ M ❜ ⇢ Atomic mass

• ❛ N ❜ ⇢ Avagadro Number (6.02 × 10²³)

___________________________