Q. If the density of KCl is 1.98 g/cm³ and its molecular weight is 74.5 g/mol, the edge length of its unit cell is approximately:

यदि KCl का घनत्व 1.98 g/cm³ है और इसका आणविक भार 74.5 g/mol है, तो इसकी इकाई कोशिका की किनारे की लंबाई लगभग है:

A
6.28 Å
B
5.28 Å
C
4.28 Å
D
3.28 Å

Explanation

Using the density formula for cubic unit cells:

ρ=Z×MNA×a3\rho = \frac{Z \times M}{N_A \times a^3}

Where:

  • ρ=1.98 g/cm3\rho = 1.98 \text{ g/cm}^3

  • M=74.5 g/molM = 74.5 \text{ g/mol}

  • NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23} \text{ mol}^{-1}

  • Z=4Z = 4 (number of formula units per unit cell in FCC structure of KCl)

  • aa = edge length in cm

Rearranging:

a3=Z×Mρ×NAa^3 = \frac{Z \times M}{\rho \times N_A}

Calculate a3a^3 and take cube root, then convert to Å (1 cm = 10810^{8} Å), which gives approximately 6.28 Å.

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