Question

Silver forms BCC lattice with edge length of 408.6 x 10-12 m. Calculate the density of

the silver. (Atomic mass of Ag =107.9 g/mol and NA= 6.022 X 1023 atoms/ mol)​

Answer 1

Explanation:

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Answer 2

Given:

• The edge length of unit cell = $408.6*10^{-12}$ pm = $408.6*10^{-10}$ cm.
• The Atomic mass of Ag = $107.9gmol^{-1}$
• Avagadro's Number = $6.022*10^{23} atoms mol^{-1}$

To Find:

• The Density of silver.

Solution:

• The volume of the unit cell = $a^{3}$ = $(408.6*10^{-10})^{3}$ = $68217388.06 * 10^{-30}$
• $a^{3} = 68.21*10^{-24} cm^{3}$  
• BCC unit cell has 2 atoms per unit cell.
• The formula to find the mass of one Ag atom is as follows,
• Mass of one silver atom = $\frac{Molar mass of Ag}{Avagadro's Number}$ = $\frac{107.9gmol^{-1} }{6.022*10^{23} mol^{-1} }$ = $17.91 * 10^{-23} g$
• We found the mass on one Ag atom above the formula of mass of unit cell requires it and the formula is,
• Mass of unit cell = $2*17.91*10^{-23}$ = $35.8*10^{-23} g$
• Density of silver = $\frac{mass of unit cell}{volume of unit cell}$ = $\frac{35.8*10^{-23}g}{68.21*10^{-24} cm^{3} }$ = $0.524*10^{1}gcm^{-3}$ = $5.2gcm^{-3}$

The Density of silver(Ag) = $5.2gcm^{-3}$