wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

The limiting molar conductivity for KCl, NaBr and NaCl are 100, 120, 130 S cm-2mol-1 respectively. Calculate the limiting molar conductivity for KBr?


A

110 S cm-2mol-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

210 S cm-2mol-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

350 S cm-2mol-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

90 S cm-2mol-1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

90 S cm-2mol-1


We know from Kohlrausch law that for

E0(KCl) = λ0Cl- + λ0K+ = 100 scm-2mol-1

E0(NaBr) = λ0Br- + λ0Na+ = 120 scm-2mol-1

E0(NaCl) = λ0Na+ + λ0Cl- = 130 scm-2mol-1

Subtracting ( iii) from ( i) and add (ii) to it

100 S cm-2mol-1 - 130 S cm-2mol-1 + 120 S cm-2mol-1 = 90 S cm-2mol-1


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electrolysis and Electrolytes_Tackle
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon