Discussion
Molar conductance of monobromoacetic acid at infinite dilution is calculated to be x × 10−4 S m2 mol−1 at 25 °C.
The value of x is (round off to the nearest integer)
Given:
| Electrolyte | Limiting molar conductance at 25 °C in 10−4 S m2 mol−1 |
|---|---|
| HBr | 427.95 |
| KBr | 151.64 |
| CH2BrCOOK | 112.72 |
Answer: C
The value of x is 389.
Kohlrausch's law of independent migration lets the limiting molar conductance of a weak acid be built from strong electrolytes. Adding HBr and CH2BrCOOK and subtracting KBr leaves exactly H+ and CH2BrCOO-:
\[ \begin{aligned} \Lambda^{0}_{\text{acid}} &= \Lambda^{0}_{HBr} + \Lambda^{0}_{CH_2BrCOOK} - \Lambda^{0}_{KBr} \\ &= 427.95 + 112.72 - 151.64 \\ &= 389.03 \end{aligned} \]
in units of 10-4 S m2 mol-1, which rounds to 389.
467 is 427.95 + 151.64 - 112.72, which swaps the roles of KBr and the potassium salt, and 164 is 427.95 - 151.64 - 112.72, which subtracts both salts. 195 is about half the correct value. Only the combination that cancels K+ and Br- leaves the acid's own ions.
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