Answer to Question #336586 in General Chemistry for Siyabonga

Question

Suppose that a solution is prepared by mixing 25 mL (0.22 mol, 16 g) of pentane (the solute) with 45 mL (0.34 mol, 30 g) of hexane (the solvent). If the volumes add on mixing,

What is the mole fraction of the pentane?

[1] 0.35

[2] 0.39

[3] 0.64

[4] 39%

[5] 64 % 

Answer 1

Hello !

Correct answer is [2] and [4] . x(pentane) = 0.22/(0.22+0.34) = 0.39 or 39%