How much heat (in kJ) is required to raise the temperature of 202.8 g of ethanol...
(specific heat = 2.46 J/g°C, ΔHvap = 48.6 kJ/mol)
Homework Answers
Given that
202.8 g of ethanol
Number of moles = 202.8 g of ethanol /46.07 g/ mole
= 4.40 moles
ΔHvap = number of moles * ΔHvap
= 4.40 mole *48.6 KJ/ mole
= 213.84 KJ
ΔH=m(c(Tb−T1)+ΔHvap)
= 202.9 g * 2.46 J/g°C (78.37 °C - 40.34 °C )+ 213.84 KJ
= 202.9 g * 2.46 J/g°C (38.03 °C )+ 213.84 KJ
= 18982.07 J +213.84 KJ
=18.982 kJ+213.84 KJ
=232.822 kJ
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