Question

Find the density of lattice, A compound AB has a rock type structure with A$:$B$=1:1$. The formula mass of AB is $6.023$ Y amu and the closed A-B distance is $Y^{1/3}$ nm.

• A. $4$ kg$/m^3$
• B. $5$ kg$/m^3$
• C. $40$ kg$/m^3$
• D. $50$ kg$/m^3$

Answer 1

The correct option is B $5$ kg$/m^3$

we have AB as a rock structure (A:B::1:1)

that is FCC structure(n=4) and formula weight of of ab is 6.023 y g closest distance $A-B=y^{\frac{1}{3}}nm$

edge length of unit cell, a= $2(A+B-)$= $2\times y^{\frac{1}{3}}\times {10}^{-9}m$

density of AB= $\frac{n\times mol.weight}{N_A\times V}$

Where $N_A$ is the Avogadro's number and V= volume of the lattice.

Density of AB= $\frac{4\times 6.023y\times {10}^{-3}}{6.023\times {10}^{23}\times {(2\times y^{\frac{1}{3}}\times {10}^{-9})}^3}kg/m^3$

= $\frac{10}{2}=5kg/m^3$