Saved Bookmarks
| 1. |
The mole fraction of benzene in a solution in toluene is 0.50. Calculate the weight percent of benzene in the solution. |
|
Answer» Solution :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.` |
|