Question

求助！大一的化学题！！！！！！！！

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？

Answer 1

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。