Question

Calculate the total quantity of heat required to convert 25.0 g of liquid ccl4(l from 35.0°c to gaseous ccl4 at 76.8°c (the normal boiling point for ccl4. the specific heat of ccl4(l is its heat of fusion is and its heat of vaporization is

Answer 1

The solution would be like this for this specific problem:

25.0 g CCl4 x (1 mole CCl4 / 153.8 g CCl4) = 0.163 moles CCl4

(29.82 kJ / mole)(0.163 moles) = 4.86 kJ

total heat = 1.11 kJ + 4.86 kJ = 5.97 kJ

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