Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9
AI Mentor
Check Your IQ
Free Expert Demo
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course
Offline Centres

Q.

If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.02 × 1023 mol-1)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

In NaCl crystal2Na+ replaced by one Sr+2;1 mole of NaCl contain 10-4100=10-6mole of Sr+2so number of cationic vacancies= 10-6 ×6.02×1023 mol-1=6.02×1017 mol-1
Watch 3-min video & get full concept clarity

courses

No courses found

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon