The enthalpies of neutralization of HCIO4 and CI3C-COOH are - 13.5 kcal/g - equivalent and- 13.5 kcal/g - equivalent respectivelywhen 40 g of solid NaOH is added to a mix of 1 g mol.HClO4 and 1 g mol CI3C-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 mix. is 6.45 kcal.

∆H for the reaction of NaOH with the mix. is 13.8 kcal.

After reaction the total no. of moles of acid remained are 0.5 (HClO4 and CI3C-COOH)

After reaction the total weight of acid remained is 147.75 g. (HClO4 + Cl3Cl-COOH) and ∆H for the reaction of NaOH with mix. is -13.8 kcal.

answer is D.

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Detailed Solution

HClO4+NaOH→NaClO4+H2OCl3C−COOH+NaOH→Cl3C−COONa+H2OLet x mol ofNaOH are used in first reaction and (1 - x) mol of NaOH are used in second reaction. Given x1−x=3⇒x=34ΔH=−13.5×34+(−14.7)×14=−40.5+14.74=−13.8kcal Weak acids =34(1+35.5+64)+334(35.5×3+24+32+1) = 147.75 g

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