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# How many unit cells are present in a cube shaped ideal crystal of NaCl of mass of 1gHow many unit cells are present in a cube shaped ideal crystal of NaCl of mass of 1g

10 months ago
Dear student

The number of Na+ ions present in one unit cell = 12×1/4 + 1 =4 atoms per unit cell.

The number of Cl- ions present in one unit cell = 6×1/2 + 8×1/8 = 4 atoms per unit cell.

So, each unit cell has 4 atoms of each Na+ and Cl-. Hence there are total 4 molecules of NaCl in each unit cell.

As the mass of -Avogadro number of molecules- of NaCl is 58.5g.

Then, number of molecules in one gram = 6.022× 10^23 ÷ 58.5 = 1.03×10^22 molecules

As we found above that each unit cell contains 4 molecules of NaCl hence we can now easily find the number of unit cells containing 1.03×10^22 molecules = 1.03×10^22 ÷ 4 = 2.575×10^21 unit cells

So, 2.575×10^21 unit cells contains 1gm of NaCl molecules.

one month ago
Dear student

The number of Na+ ions present in one unit cell = 12×1/4 + 1 =4 atoms per unit cell.

The number of Cl- ions present in one unit cell = 6×1/2 + 8×1/8 = 4 atoms per unit cell.

So, each unit cell has 4 atoms of each Na+ and Cl-. Hence there are total 4 molecules of NaCl in each unit cell.

As the mass of -Avogadro number of molecules- of NaCl is 58.5g.

Then, number of molecules in one gram = 6.022× 10^23 ÷ 58.5 = 1.03×10^22 molecules

As we found above that each unit cell contains 4 molecules of NaCl hence we can now easily find the number of unit cells containing 1.03×10^22 molecules = 1.03×10^22 ÷ 4 = 2.575×10^21 unit cells

So, 2.575×10^21 unit cells contains 1gm of NaCl molecules.

one month ago
Dear student

The number of Na+ ions present in one unit cell = 12×1/4 + 1 =4 atoms per unit cell.

The number of Cl- ions present in one unit cell = 6×1/2 + 8×1/8 = 4 atoms per unit cell.

So, each unit cell has 4 atoms of each Na+ and Cl-. Hence there are total 4 molecules of NaCl in each unit cell.

As the mass of -Avogadro number of molecules- of NaCl is 58.5g.

Then, number of molecules in one gram = 6.022× 10^23 ÷ 58.5 = 1.03×10^22 molecules

As we found above that each unit cell contains 4 molecules of NaCl hence we can now easily find the number of unit cells containing 1.03×10^22 molecules = 1.03×10^22 ÷ 4 = 2.575×10^21 unit cells

So, 2.575×10^21 unit cells contains 1gm of NaCl molecules.