The enthalpies of neutralization of ${HClO}_{4} and {Cl}_{3} C - COOH$ are
                                      -13.5 kcal/g-equivalent and
                                      -13.5 kcal/g-equivalent respectively
          When 40g of solid NaOH is added to a mixtureof 1g mol. ${HClO}_{4}$ and 1g mol. ${Cl}_{3} C - COOH$ , sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3:1 then

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$∆ H$ For the reaction of NaOH with the mixture is 6.45 kcal.

$∆ H$ For the reaction of NaOH with the mixture is 13.8 kcal

After reaction the total number of moles of acid remained are 0.5 ( ${HClO}_{4}$ and ${Cl}_{3} C - COOH$ )

After reaction the total weight of acid remained is 147.75gm. $({HClO}_{4} + {Cl}_{3} C - COOH)$ and $∆ H$ for the reaction of NaOH with mixture is -13.8 kcal.

answer is D.

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Detailed Solution

${HClO}_{4} + NaOH \to {NaClO}_{4} + H_{2} O {Cl}_{3} C - COOH + NaOH \to {Cl}_{3} C - COONa + H_{2} O$

Let x mol of NaOH are used in 1st reaction and (1-x) mol of NaOH are used in 2nd reaction given $\frac{x}{1 - x} = 3 \Rightarrow x = \frac{3}{4}$

$ΔH = - 13.5 \times \frac{3}{4} + (- 14.7) \times \frac{1}{4} = \frac{- 40.5 - 14.7}{4} = 13.8 kcal$

$Weak acids = \frac{1}{4} (1 + 35.5 + 64) + \frac{3}{4} (35.5 \times 3 + 24 + 32 + 1) = 147.75 gm$

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