Question

A solid having density of 9 × 10³ kg m^-3 forms face centered cubic crystals of edge length \(200\sqrt 2\) pm. What is the molar mass of the solid? 

[Avogadro constant = 6 × 10²³ mol^-1, π = 3]
1. 0.03050 kg mol^-1
2. 0.4320 kg mol^-1
3. 0.0432 kg mol^-1
4. 0.0216 kgmol^-1

Answer 1

Correct Answer - Option 1 : 0.03050 kg mol^-1

Concept:

General expression for density of unit cell:

Unit Cell:

The smallest group of atoms which has the overall symmetry of a crystal, and from which the entire lattice can be built up by repetition in three dimensions is termed as Unit Cell.

The density of unit cell is given by the following

\({\rm{Density\;of\;unit\;cell}} = \frac{{{\rm{mass\;of\;unit\;cell}}}}{{{\rm{volume\;ofunit\;cell}}}}\)

\({\rm{Density\;of\;unit\;cell}} = \frac{{\rm{m}}}{{\rm{V}}} = {\rm{z\times}}\frac{{\rm{m}}}{{{{\rm{a}}^3}}}\)      

\(d = \frac{{M \times Z}}{{{N_A} \times {a^3}\;}} \Rightarrow M = \frac{{d \times {N_A} \times {a^3}}}{Z}\;\)

Where

M = molar mass

Calculation:

Density of a crystal 

\(d = \frac{{M \times Z}}{{{N_A} \times {a^3}\;}} \Rightarrow M = \frac{{d \times {N_A} \times {a^3}}}{Z}\;\)

Given, d = 9 × 10³ kg m^-3

M = Molar mass of the solid

Z = 4 (for fcc crystal)

N_A = Avogadro's constant = 6 × 10²³ mol^-1

a = Edge length of the unit cell 

\(= 200\sqrt 2 {\rm{\;pm}} = 200\sqrt 2 \times {10^{ - 12}}{\rm{\;m}}\)

On substituting all the given values, we get

= (9 × 10³) kg m^-3 × (6 × 10²³) mol^-1

\(= \frac{{{{\left( {200\sqrt 2 \times {{10}^{ - 12}}} \right)}^3}{m^3}}}{4}\)

= 0.0305 kg mol^-1