A metal has a fcc lattice. The edge length of the unit cell is 404 pm . The density of the metal is 2.72 ~g ~cm^ -3 . The molar mass of the metal is ( N_A Avogadro's constant =6.02 × 10^ 23 ~mol^ -1 )

Correct answer: C. 27 ~g ~mol^ -1 Given, cell is fcc, so Z=4 Edge length, a=404 pm=4.04 × 10^ -8 ~cm Density of metal, d=2.72 ~g ~cm^ -3 N_A=6.02 × 10^ 23 ~mol^ -1 Molar mass of the metal, M= ? We know that density, d=(Z × M)/(2^3 · N_A) aligned & =(d · a^3 · N_A)/(Z) & = 2.72 ×(4.04 × 10^ -8 )^3 × 6.02 × 10^ 23 4 & =27 ~g ~mol^ -1 aligned

NEET 2013

Exam: NEET
Year: 2013
Subject: Chemistry
Chapter: Solid State

NEET 2013 · Chemistry · Solid State

A metal has a fcc lattice. The edge length of the unit cell is 404...

A metal has a fcc lattice. The edge length of the unit cell is

404 \mathrm{pm}

. The density of the metal is

2.72 \mathrm{~g} \mathrm{~cm}^{-3}

. The molar mass of the metal is (

N_A

Avogadro's constant

=6.02 \times 10^{23} \mathrm{~mol}^{-1}

)

Options

A $40 \mathrm{~g} \mathrm{~mol}^{-1}$
B $30 \mathrm{~g} \mathrm{~mol}^{-1}$
C $27 \mathrm{~g} \mathrm{~mol}^{-1}$
D $20 \mathrm{~g} \mathrm{~mol}^{-1}$

Correct answer

C. $27 \mathrm{~g} \mathrm{~mol}^{-1}$

Step-by-step solution

Given, cell is fcc, so

Z=4

Edge length,

a=404 \mathrm{pm}=4.04 \times 10^{-8} \mathrm{~cm}

Density of metal,

d=2.72 \mathrm{~g} \mathrm{~cm}^{-3}N_A=6.02 \times 10^{23} \mathrm{~mol}^{-1}

Molar mass of the metal,

M=

? We know that density,

d=\frac{Z \times M}{2^3 \cdot N_A}

\begin{aligned} \thereforeM & =\frac{d \cdot a^3 \cdot N_A}{Z}\\ & =\frac{2.72 \times\left(4.04 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}}{4}\\ & =27 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}

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