Verified answer

Answer

Part1

By Clausius-Clapeyron equation

logP2/P1 = delta H/(2.303*R) * (1/T1-1/T2)

P1=92 torr P2=372, T1=296k, T2=318

deltaH = heat of vaporisation in cal/mole

T2 and T1 are temp in kelvin

R=1.99 cal/mole-k

log372/92=0.6067

2.303*R=2.303*1.99=4.5829cal/mole-k

1/T1-1/T2=0.0002337K

Therefore, deltaH=0.6067*4.5829/0.0002337 =11897 cal or 11897/239kj=49.78kj

Part2

Let normal boiling point is T2

Corresponding vap pressure P2 would be atm pr =760 torr

log760/92=0.9170 , P1=92 torr

2.303*R=2.303*1.99=4.5829cal/mole-k

1/T1-1/T2= (1/296-1/T2) , T1=273+23=296k

1/T1-1/T2=( 0.003378 - 1/T2)

Heat of vaporisation = 47.9kj given=47.9*239cal = 11448cal 1 kJ = 239.0057

Therefore, 0.9170*4.5829/11448=0.003378-1/T2

0.0003671=0.003378-1/T2

1/T2=0.003378-0.0003871=0.0030109

T2=332K or 59 degc

I did it myself and it turns out the Answer above is actually the correct answer (both parts) however sapling chooses to be a dick in require that you enter 4 significant figures so therefore your answer will be 49.750