wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

0.02 mole of [Co(NH3)5Br]Cl2 and 0.02 mole of [Co(NH3)5Cl]SO4 are present in 200 cc of a solution X. The number of moles of the preceipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively :

A
0.02,0.02
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.01,0.02
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.02,0.04
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.04,0.02
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 0.04,0.02
0.02 mole of [Co(NH3)5Br]Cl2 will dissociate to give 2×0.02=0.04 moles of chloride ions. On reaction with silver nitrate, 0.04 moles of silver chloride precipitate will be obtained.
[Co(NH3)5Br]Cl2[Co(NH3)5Br]2++2Cl
0.02 mole of [Co(NH3)5Cl]SO4 will dissociate to give 0.02 moles of sulphate ions. On reaction with barium sulphate, 0.02 moles of barium sulphate precipitate will be obtained.
[Co(NH3)5Cl]SO4[Co(NH3)5Cl]2++SO24

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Werner's Coordination Theory
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon