SATHEE: The Solid State - Result Question 15

The Solid State - Result Question 15

####15. A metal has an $f c c$ lattice. The edge length of the unit cell is $404 p m$ . The density of the metal is $2.72 g c m^{- 3}$ . The molar mass of the metal is : $(N_{A} .$ , Avogadro’s constant $. = 6.02 \times 10^{23} m o l^{- 1})$

(a) $30 g m o l^{- 1}$

(b) $27 g m o l^{- 1}$

(d) $40 g m o l^{- 1}$

[NEET 2013]

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Solution:

(b) Density is given by

$d = \frac{Z \times M}{N_{A} a^{3}}$ ; where $Z =$ number of formula units present in unit cell, which is 4 for $f c c$

$a =$ edge length of unit cell. $M =$ Molecular mass

$2.72 = \frac{4 \times M}{6.02 \times 10^{23} \times (404 \times 10^{- 10})^{3}}$

$(∵ 1 p m = 10^{- 10} c m)$

$M = \frac{2.72 \times 6.02 \times (404)^{3}}{4 \times 10^{7}} = 26.99 = 27 g m o l^{- 1}$

The Solid State - Result Question 15

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The Solid State - Result Question 15

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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