6. ⇒ (MHT CET 2023 11th May Morning Shift )

Which formula is used to calculate edge length in bcc structure?

A. $a = \frac{\sqrt{3} r}{4}$

B. $a = \frac{4}{\sqrt{3} r}$

C. $a = \frac{\sqrt{3}}{4 r}$

D. $a = \frac{4 r}{\sqrt{3}}$

Correct Option is (D)

Currently no explanation available

7. ⇒ (MHT CET 2023 11th May Morning Shift )

What is atomic mass of an element with $BCC$ structure and density $10 g {cm}^{- 3}$ having edge length $300 pm$ ?

A. $51.0 g {mol}^{- 1}$

B. $60.0 g {mol}^{- 1}$

C. $81.3 g {mol}^{- 1}$

D. $96.8 g {mol}^{- 1}$

Correct Option is (C)

For bcc unit cell, $n = 2$ .

$\begin{aligned} Density of bcc unit cell = ρ = \frac{Mn}{a^{3} N_{A}} \\ M = \frac{ρ \times a^{3} \times N_{A}}{n} \\ M = \frac{10 g {cm}^{- 3} \times {(3 \times 10^{- 8} cm)}^{3} \times 6.022 \times 10^{23} atom {mol}^{- 1}}{2 atoms} \\ M = 81.3 g {mol}^{- 1} \end{aligned}$

8. ⇒ (MHT CET 2023 11th May Morning Shift )

What is the number of unit cells present in a cubic pack crystal lattice having 4 atoms per unit cell and weighing $0.60 g$ (molar mass $60 g {mol}^{- 1})$ ?

A. $1 \times 10^{21}$

B. $1.5 \times 10^{21}$

C. $3.0 \times 10^{21}$

D. $6.0 \times 10^{21}$

Correct Option is (B)

Number of moles $= \frac{0.6 g}{60 g {mol}^{- 1}} = 0.01 mol$

Total number of atoms $= 0.01 \times 6.022 \times 10^{23}$

Number of atoms per unit cell $= 4$ (Given)

$∴$ Number of unit cells present in the given cubic crystal lattice

$\begin{aligned} = \frac{0.01 \times 6.022 \times 10^{23}}{4} \\ = 1.5 \times 10^{21} unit cells \end{aligned}$

9. ⇒ (MHT CET 2023 10th May Evening Shift )

Calculate the number of atoms present in unit cell of an element having molar mass $190 g {mol}^{- 1}$ and density $20 g {cm}^{- 3}$ .

$[a^{3} \cdot N_{A} = 38 {cm}^{3} {mol}^{- 1}]$

A. 1

B. 2

C. 6

D. 4

Correct Option is (D)

$\begin{aligned} Density (ρ) = \frac{M n}{a^{3} N_{A}} \\ 20 = \frac{190 \times n}{38} \\ n = \frac{20 \times 38}{190} = 4 \end{aligned}$

10. ⇒ (MHT CET 2023 10th May Morning Shift )

Calculate the molar mass of an element with density $2.7 g {cm}^{- 3}$ that forms fcc structure. $[a^{3} \cdot N_{A} = 40 {cm}^{3} {mol}^{- 1}]$

A. $112 g {mol}^{- 1}$

B. $54 g {mol}^{- 1}$

C. $27 g {mol}^{- 1}$

D. $78 g {mol}^{- 1}$

Correct Option is (C)

For fcc unit cell, $n = 4$ .

$\begin{aligned} Density (ρ) = \frac{M n}{a^{3} N_{A}} \\ 2.7 = \frac{M \times 4}{40} \\ M = \frac{2.7 \times 40}{4} = 27 g {mol}^{- 1} \end{aligned}$

⇒Solid State