1. ⇒ ( MHT CET 2023 12th May Evening Shift)

Calculate the density of an element having molar mass $27 g {mol}^{- 1}$ that forms fcc unit cell. $[a^{3} \cdot N_{A} = 38.5 {cm}^{3} {mol}^{- 1}]$

A. $2.8 g {cm}^{- 3}$

B. $3.5 g {cm}^{- 3}$

C. $2.1 g {cm}^{- 3}$

D. $4.1 g {cm}^{- 3}$

Correct Option is (A)

$\begin{aligned} Density, ρ = \frac{M n}{a^{3} N_{A}} \\ ρ = \frac{27 g {mol}^{- 1} \times 4 atom}{38.5 {cm}^{3} {mol}^{- 1}} \\ = 2.8 g {cm}^{- 3} \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Evening Shift )

Calculate the edge length of simple cubic unit cell if radius of an atom is $167.3 pm$ .

A. 473.2 pm

B. 334.6 pm

C. 386.3 pm

D. 836.5 pm

Correct Option is (B)

For simple cubic unit cell, $a = 2 r = 2 \times 167.3 pm = 334.6 pm$

For simple cubic unit cell, edge length is twice the radius of an atom.

3. ⇒ (MHT CET 2023 12th May Morning Shift )

Calculate the molar mass of an element having density $7.8 g {cm}^{- 3}$ that forms bcc unit cell. $[a^{3} \cdot N_{A} = 16.2 {cm}^{3} {mol}^{- 1}]$

A. $63.18 g {mol}^{- 1}$

B. $61.23 g {mol}^{- 1}$

C. $59.31 g {mol}^{- 1}$

D. $65.61 g {mol}^{- 1}$

Correct Option is (A)

For bcc unit cell, $n = 2$ .

$\begin{aligned} Density (ρ) = \frac{M n}{a^{3} N_{A}} \\ 7.8 = \frac{M \times 2}{16.2} \\ M = \frac{7.8 \times 16.2}{2} = 63.18 g {mol}^{- 1} \end{aligned}$

4. ⇒ (MHT CET 2023 11th May Evening Shift )

Calculate the molar mass of metal having density $9.3 g {cm}^{- 3}$ that forms simple cubic unit cell. $[a^{3} \cdot N_{A} = 22.6 {cm}^{3} {mol}^{- 1}]$

A. $210.2 g {mol}^{- 1}$

B. $105.3 g {mol}^{- 1}$

C. $52.6 g {mol}^{- 1}$

D. $70.2 g {mol}^{- 1}$

Correct Option is (A)

For simple cubic unit cell, $n = 1$ .

$\begin{aligned} Density (ρ) = \frac{M n}{a^{3} N_{A}} \\ 9.3 = \frac{M \times 1}{22.6} \\ M = \frac{9.3 \times 22.6}{1} = 210.2 g {mol}^{- 1} \end{aligned}$

5. ⇒ (MHT CET 2023 11th May Evening Shift )

Find the radius of an atom in fcc unit cell having edge length

A. 202.5 pm

B. 175.3 pm

C. 143.2 pm

D. 181.0 pm

Correct Option is (C)

For fcc crystal structure, $r = \frac{\sqrt{2}}{4} a$

$∴ r = \frac{1.414 \times 405}{4} = 143.2 pm$

⇒Solid State