MHT CET 202621 AprilMorning Shift
- Exam
- MHT CET
- Year
- 2026
- Date
- 21 April
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Electrochemistry
MHT CET 2026 · 21 April · Morning Shift · Chemistry · Electrochemistry
The molar conductance of NaCl, HCl and CH_3COONa at infinite...
The molar conductance of NaCl, HCl and at infinite dilution are 126.45, 426.16 and 91.0 S respectively. Calculate the molar conductance of at infinite dilution?
Options
- A540.48 S
- B201.28 S
- C390.71 S
- D698.28 S
Correct answer
C. 390.71 S
Step-by-step solution
According to Kohlrausch's law of independent migration of ions:
\Lambda^{}^{\circ}_m(\text{CH}_3\text{COOH}) = \Lambda^{}^{\circ}_m(\text{CH}_3\text{COONa}) + \Lambda^{}^{\circ}_m(\text{HCl}) - \Lambda^{}^{\circ}_m(\text{NaCl})
Substituting the given values:
\Lambda^{}^{\circ}_m(\text{CH}_3\text{COOH}) = 91.0 + 426.16 - 126.45
\Lambda^{}^{\circ}_m(\text{CH}_3\text{COOH}) = 517.16 - 126.45 = 390.71 \text{ S cm}^2 \text{ mol}^{-1}
Answer: 390.71 S
\Lambda^{}^{\circ}_m(\text{CH}_3\text{COOH}) = \Lambda^{}^{\circ}_m(\text{CH}_3\text{COONa}) + \Lambda^{}^{\circ}_m(\text{HCl}) - \Lambda^{}^{\circ}_m(\text{NaCl})
Substituting the given values:
\Lambda^{}^{\circ}_m(\text{CH}_3\text{COOH}) = 91.0 + 426.16 - 126.45
\Lambda^{}^{\circ}_m(\text{CH}_3\text{COOH}) = 517.16 - 126.45 = 390.71 \text{ S cm}^2 \text{ mol}^{-1}
Answer: 390.71 S