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CCl₂F₂ cooled 1.25g sample at constant atmospheric pressure of 1 atm from 320 K to 293 K. During cooling the sample volume decreased from 274 to 248 mL. Calculate $∆$ H for the CCl₂F₂ for this process. (C_p = 80.7 J mol^–1 K^–1)

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–19.83 J

22.51 J

–22.51 J

19.88 J

answer is D.

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Detailed Solution

$∆ H = nCp ∆ T$

$∆ T$ = 293 - 320 = - 27 K

$∆ H = \frac{1.25 \times 80.7 \times (- 27)}{121} = - 22.51 J$

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