# The number of atoms in 100 g of an fcc crystal with density $\mathrm{d}=10\mathrm{g}/{\mathrm{cm}}^{3}$ and cell edge equal to 100 pm is equal to

1. A

$4×{10}^{25}$

2. B

$3×{10}^{25}$

3. C

$2×{10}^{25}$

4. D

$1×{10}^{25}$

FREE Lve Classes, PDFs, Solved Questions, PYQ's, Mock Tests, Practice Tests, and Test Series!

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.

### Solution:

$\frac{\mathrm{Z}×\mathrm{M}}{{\mathrm{a}}^{3}×\mathrm{N}}$

$\mathrm{M}=\frac{\mathrm{\rho }×{\mathrm{a}}^{3}×{\mathrm{N}}_{0}×{10}^{-30}}{\mathrm{z}}$

$=\frac{10×{\left(100\right)}^{3}×\left(6.02×{10}^{23}\right)×{10}^{-30}}{4}=1.505$gr

Number of atoms in $100\mathrm{g}=\frac{6.02×{10}^{23}}{1.505}×100=4×{10}^{25}$

## Related content Join Infinity Learn Regular Class Program!

Sign up & Get instant access to FREE PDF's, solved questions, Previous Year Papers, Quizzes and Puzzles!

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.