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Answer 1 · 2016-07-08T17:12:07

#~~0.26#

Let in the mixture

#"No. of moles of " ""^10B=x " mol"#

#"No. of moles of " ""^11B=y " mol"#

#:. "Mole fraction of "^10B=x/(x+y)#

So

#x xx 10.012+ yxx11.093=(x+y)xx10.811#

#=> yxx11.093-yxx10.811=x xx10.811-x xx 10.012#

#=>0.282y=0.799x#

#=>y/x=799/282#

#=>y/x+1=799/282+1#

#=>(x+y)/x=(282+799)/282=1081/282#

#=>x/(x+y)=282/1081~~0.26#

#:. "Mole fraction of "^10B=x/(x+y)=282/1081~~0.26#