Question

What is the number of unit cells in 936 amu of NaCl?​

Answer 1

as you know that in NaCl, Cl ions forms Fcc and Na ions occupy octahedral voids.

The formula unit cell is 4 i.e. 1 unit cell of NaCl contains 4 NaCl particles.

58.5 gm of NaCl contains-6.022*10^23 atoms

4 atoms constitute 1 unit cell

therefore, 6.022*10^23 atoms constitute 6.022*10^23/4

= 1.5 * 10^23

Answer 2

Answer:

The number of the unit cells of NaCl  in 936a.m.u. will be equal to four.

Explanation:

• We know that in NaCl unit cell Na⁺ is present at octahedral voids and Cl⁻ is present in FCC arrangement. Therefore Na = 4 and Cl =4. Four NaCl are present in one unit cell.
• Mass of one NaCl unit $= 23+35.5=58.5 amu$

        Mass of four NaCl units$=4*58.5=234amu$

• 234a.m.u. has no. of NaCl unit cell =1

        936a.m.u. has  number of NaCl unit cells $=\frac{936}{234} =4$ unit cell