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CHCl3在40℃时蒸汽压为49.3kPa.于此温度和101.3kPa压力下,有4.00L空气缓慢的通过CHCl3(即每个气泡都为CHCl3蒸汽所饱和)求:(1)空气和CHCl3混合气体的体积是多少?(2)被空气带走的CHCl3质量是多少
题目详情
CHCl3在40℃时蒸汽压为49.3kPa.于此温度和101.3kPa压力下,有4.00L空气缓慢的通过CHCl3 (即每个气泡都为CHCl3蒸汽所饱和)求:(1)空气和CHCl3混合气体的体积是多少?(2)被空气带走的CHCl3质量是多少?
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答案和解析
CHCl3的molar density是partial pressure divided by total pressure, which is 49.3/101.3, 4 liters air in normal condition gives 4/25.69mol air, (PV=NRT,T=313K,P=101300Pa)
therefore, we have the equation x/(x+4/25.69)=49.3/101.3
by compution we get x=0.14763mol, which is the mole of CHCl3.
(1)Thus the misture volumn should be approximately 1000*0.14763*8.314*313/101300+4=7.8L
(2)the mass should be 0.14763mol*119.5=17.64g
习惯用英文,希望你看得懂.
therefore, we have the equation x/(x+4/25.69)=49.3/101.3
by compution we get x=0.14763mol, which is the mole of CHCl3.
(1)Thus the misture volumn should be approximately 1000*0.14763*8.314*313/101300+4=7.8L
(2)the mass should be 0.14763mol*119.5=17.64g
习惯用英文,希望你看得懂.
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