An element crystallizes in fcc lattice with cell edge of 300 pm. The density of element is 10.8gcm^{-3}. Calculate the number of atoms in 108g of the element.

 

 

 

 
 
 
 
 

Answers (1)

Given:

FCC lattice, z=4

\rho = 10.8gcm^{-3}

a=300pm = 300\times 10^{-10}cm

\rho =\frac{z\times M}{a^{3}\times N_{A}}

M = \frac{\rho \times a^{3\times N_{A}}}{z}

        = \frac{10.8\times \left ( 300\times 10^{-10} \right )\times 6.022\times 10^{23}}{4}=175.6g

n = \frac{108}{175.6}=0.615mol

1 mol contains → N_{A} atoms

0.615 mol → 0.615\times N_{A} = 14.81 \times 10^{23}atoms

 

Most Viewed Questions

Related Chapters

Preparation Products

Knockout CUET (Physics, Chemistry and Mathematics)

Complete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout CUET (Physics, Chemistry and Biology)

Complete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Knockout NEET (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions