某水溶液的凝固点是272.15K,试计算:(1)此溶液的沸点。(2)298.15K时的渗透压。已知Kf(H2O)=1.86℃.kg.mol-1,Kb(H2O)=0.52℃.kg.mol-1,纯水在正常气压时的沸点是373.15K。
正确答案:
(1)△Tf=Kf*bB,bB=△Tf/Kf
△Tb=Kb*bB=Kb*△Tf/Kf=0.52*(273.15-272.15)/1.86=0.28
Tb=373.15+0.28=373.43k
(2)bB=nB/mA=nB/(P水V),CB=nB/V
∏=CBRT=(nB/V)RT=P水bBRT
=l*[(273.15-272.15)/1.86]*8.314*103*298.15
=1.333*103Kpa
△Tb=Kb*bB=Kb*△Tf/Kf=0.52*(273.15-272.15)/1.86=0.28
Tb=373.15+0.28=373.43k
(2)bB=nB/mA=nB/(P水V),CB=nB/V
∏=CBRT=(nB/V)RT=P水bBRT
=l*[(273.15-272.15)/1.86]*8.314*103*298.15
=1.333*103Kpa
答案解析:有
微信扫一扫手机做题