The edge length of NaCl unit cell is 564 pm. What is the density of NaCl in g/`cm^(3)`?

Options

A. `1.082 g cm^(-3)`
B. `2.165 m cm^(-3)`
C. `3.247 g cm^(-3)`
D. `4.330 g cm^(-3)`

Answer

B

Explanation

In sodium chloride unit cell, there are four `Na^(+)` and four `Cl^(-)` ions. Hence
density = `("Mass of atoms in a unit cell")/("Volume of a unit cell ")`
`=(4(23+35.5) g "mol"^(-1))/((6.023xx10^(23) "moll"^(-1)) xx (564xx10^(=10) cm)^(3))`
`=2.165 g cm^(-3)`