CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Relative decrease in vapour pressure of an aqueous solution containing 2 moles [Cu(NH3)3Cl]Cl in 3 moles H2O is 0.50. On reaction with AgNO3, this solution will form:

A
1 mol AgCl
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0.25 mol AgCl
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2 mol AgCl
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.40 mol AgCl
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1 mol AgCl
Relative lowering in vapour pressure= 0.50
ix=0.5
x= mole fraction
As, x=25=molesofsolutemolesofsolution
i=0.5×52=1.25
Now, degree of dissociation
α=i1n1=1.25121=0.25
(n=2ions=2)
So, moles of chlorine react with AgNO3=1 and Moles of AgCl precipitate= 1×0.25=0.25mol

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Types of Redox Reactions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon