The mole fraction fo benzene in a solution with toluene is 0.50 . Calculate the mass present of benzene in the solution.

Explanation

Suppose the weight percent of benzene in the solution = x. This means that in 100 g solution,
`"Mass of benzene = x g , Mass of toluene "=(100-x)g`
Mol. Mass of toluene `(C_(6)H_(5)CH_(3))=92`
`therefore" "(n_(B))/((n_(B)+n_(T)))=0.50, i.e., (x//78)/(x//78+(100-x)//92)=0.50`
`(x)/(78)xx(78xx92)/(92x+78(100-x))=0.50 or 92x=46x+3900-39x or 85x=3900 or x = 45.9%`.
Alternatively, `(n_(B))/(n_(B)+n_(T))=0.5, i.e., (w_(B)//78)/(w_(B)//78+w_(T)//92)=0.5 or (w_(B))/(78)=0.5or(w_(B))/(78)0.5((w_(B))/(78)+(w_(T))/(92))=(1)/(2)((w_(B))/(78)+(w_(T))/(92))`
`"or "(2w_(B))/(78)=(w_(B))/(78)+(w_(T))/(92)or (w_(B))/(78)=(w_(T))/(92) or (w_(T))/(w_(B))=(92)/(78) or 1+(92)/(78)`
`"or "(w_(B)+w_(T))/(w_(B))=(170)/(78)" or "(w_(B))/(w_(B)+w_(T))=(78)/(170)=0.459.` Hence, `wt%=45.9.`