Question

$A°$ $m$ for $NaCl$, $HCl$ and $NaAc$ are $126.4, 425.9$ and $91.0$ $S$ $cm^2$ $mol^{-1 }$respectively. Calculate $A°$ for $HAc$

• A. $390.5$ $S\; cm^2 \;mol^{-1}$
• B. $643.3$ $S$ $cm^2$ $mol^{-1}$
• C. $461.3$ $S$ $cm^2$ $mol^{-1}$
• D. $208.5$ $S$ $cm^2$ $mol^{-1}$

Answer 1

Answer: (A)

$390.5$ $S\; cm^2 \;mol^{-1}$

Answer 2

=$∧\degree m(HCl)+∧\degree m(NaAc)−A\degree m(NaCl)$

=(425.9+91.0−126.4)$$S $cm^2mol^{−1}$

=$390.5$ $S$ $cm^2$ $mol^{−1}$

Hence, the correct option is (A): $390.5$ $S$ $cm^2$ $mol^{−1}$