CBSE (Science) Class 12CBSE
Account
It's free!

User


Login
Create free account


      Forgot password?
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution - If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies? - CBSE (Science) Class 12 - Chemistry

Question

If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies?

Solution

It is given that NaCl is doped with 10−3 mol% of SrCl2.

This means that 100 mol of NaCl is doped with 10−3 mol of SrCl2.

Therefore, 1 mol of NaCl is doped with `10^(-3)/100` mol of SrCl2

= 10−5 mol of SrCl2

Cation vacancies produced by one Sr2+ ion = 1

Concentration of the cation vacancies Produced by 10-5 mol of Sr2+ ions = `10^(-5)xx6.022xx10^(23)`

=6.022 x 1018 mol-1

Hence, the concentration of cation vacancies created by SrCl2 is 6.022 × 108 per mol of NaCl.

Is there an error in this question or solution?

APPEARS IN

 NCERT Chemistry Textbook for Class 12 Part 1 (with solutions)
Chapter 1: The Solid State
Q: 25 | Page no. 32

Reference Material

Solution for question: If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies? concept: null - Types of Point Defects - Non-stoichiometric Defects. For the courses CBSE (Science), CBSE (Arts), PUC Karnataka Science, CBSE (Commerce)
S