Kf=20.2
苯的物质的量是n=m/M=15.6/78=0.2mol
400g=0.4kg
苯的质量浓度是b=n/m(环己烷)=0.2/0.4=0.5mol/kg
△Tf=b*Kf
10.1=0.5*Kf
Kf=10.1/0.5=20.2
将15.6g苯溶于400g环已烷中,该溶液的凝固点比纯溶剂降低10.1℃,计算环已烷的凝固
Kf=20.2
苯的物质的量是n=m/M=15.6/78=0.2mol
400g=0.4kg
苯的质量浓度是b=n/m(环己烷)=0.2/0.4=0.5mol/kg
△Tf=b*Kf
10.1=0.5*Kf
Kf=10.1/0.5=20.2