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

50 mL of 0.1(M) BaCl2 solution and 150 mL 0.1 (M) AgCl solution are mixed. Calculate the molarity of Cl ions in the final solution.

A
0.25
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.125
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 0.125
BaCl2BaCl2Ba+2+2Cl
50ml,0.1M 0.1M 0.1M 0.2M
Molarity=nVolm
AgClAgClAg++Cl
150ml,0.1M 0.1M 0.1M 0.1M
BaCl22Cl
No. of moles=MxV
=0.2x50
=10
AgCl1Cl
No. of moles=MxV
=0.2x150
=15
Total no. moles of Cl=10+15=25
Total volume=200
Molarity=25200=18=0.125
The molarity of Cl ions in the final solution is 0.125M

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative Lowering of Vapour Pressure
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon