A metal crystallizes in FCC crystal lattice with density 10 g/cm^3 . The edge length of unit cell is 100 pm. The number of unit cell present in 100 g of metal is-

Correct answer: D. 1 × 10^ 25 Volume of the metal = Mass Density = (100)/(10) = 10 cm^3 Edge length of unit cell, a = 100 pm = 10^ -8 cm Volume of one unit cell = a^3 = (10^ -8 )^3 = 10^ -24 cm^3 Number of unit cells = Total volume Volume of one unit cell = 10 10^ -24 = 1 × 10^ 25 Answer: 1 × 10^ 25

MHT CET 202621 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 21 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Solid State

MHT CET 2026 · 21 April · Morning Shift · Chemistry · Solid State

A metal crystallizes in FCC crystal lattice with density 10 g/cm^3...

A metal crystallizes in FCC crystal lattice with density 10

\text{g/cm}^3

. The edge length of unit cell is 100 pm. The number of unit cell present in 100 g of metal is-

Options

A $1 \times 10^{22}$
B $1 \times 10^{23}$
C $1 \times 10^{24}$
D $1 \times 10^{25}$

Correct answer

D. $1 \times 10^{25}$

Step-by-step solution

Volume of the metal =

\dfrac{\text{Mass}}{\text{Density}} = \dfrac{100}{10} = 10 \text{ cm}^3

Edge length of unit cell,

a = 100 \text{ pm} = 10^{-8} \text{ cm}

Volume of one unit cell =

a^3 = (10^{-8})^3 = 10^{-24} \text{ cm}^3

Number of unit cells =

\dfrac{\text{Total volume}}{\text{Volume of one unit cell}} = \dfrac{10}{10^{-24}} = 1 \times 10^{25}

Answer:

1 \times 10^{25}

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