Question

A metallic element has simple cubic arrangement where edge length = 288 pm and density = $7.5gc{m}^{-3}$. The number of atoms in 100 g of the metal are $5.58\times {10}^{23}$.

If the given statement is true enter 1,else enter 0.

Answer 1

$P=\frac{ZM}{N_0{a}^3},P=7.5ℊ/㎤;a=288pm;a=288\times {10}^{-12}m$

Simble cubic arrangement, $Z=1$

$M=\frac{P\times N_0\times a^3}{Z}=0.00010789\times {10}^6ℊ$

Number of atoms in $100ℊ$ of metal$=\frac{6.023\times {10}^{23}\times 100}{{10}^6\times 0.00010789ℊ}$

$=5.58\times {10}^{23}$

Ans: 1