Home
Class 12
CHEMISTRY
I mol of NaCl is doped with 10^(-5) mole...

I mol of NaCl is doped with `10^(-5)` mole of `SrCl_(2)` .The number of cationic vacancies in the crystal lattice will be

A

`6.022 xx 10^18`

B

`6.022 xx 10^23`

C

`6.022 xx 10^15`

D

`12.044 xx 10^20`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

If NaCl is doped with 10^(-2) mol% SrCl_(2) , what is the concentration of the cation vacancies?

If NaCl is doped with 10^(-3) mol% GaCl_(3) , what is the concentration of the cation vacancies?

If NaCl is doped with 10^(-4) mol % of SrCl_(2) , the concentration of cation vacancies will be ( N_(A) = 6.022 xx 10^(23) mol^(-1))

If NaCl is doped with 10^(-3) mole percent SrCl_(2) , what will be the concentration of cation vacancies ? ( N_(A) = 6.02 xx 10^(23) mol^(-1) )

0.5 mol of BaCl_2 is mixed with 0.2 mol of Na_3PO_4 . The maximum number of mol of Ba_3(PO_4)_2 that can be formed is :

The number of moles of NaCl in 2L of 3M NaCl solution is :

Assertion (A) : The density of crystal having Schottky defect is lowered. Reason (R ) : The crystals suffering from Schottky defect have same number of cation and anions missing from their normal lattice sites.

If 0.5 mol of CaBr_(2) is mixed with 0.2 mol of K_(3) PO_(4) , the maximum nubmer of moles of Ca_(3) (PO_(4))_(2) that can be formed is: a. 0.1 b. 0.2 c. 0.5 d.0.7