The mole fraction of benzene in a solution containing 39 % by mass in an organic solvent of molecular mass 122 is

Step-by-step solution

Given, 39 % by mass of organic solvent in benzene 39 ~g of organic solvent in 100 ~g solution. or 39 ~g of organic solvent in 100-39=61 g benzene. Now, molar mass of organic solvent =122 ~g / mol Molar mass of benzene ( C ₆ H ₆ )=12 6+6=78 ~g / mol Moles of benzene (n_ B )= 61 78 Moles of organic solvent (n_ O )= 39 122 Mole fraction of benzene = n_ B n_ B +n_ O = 61 78 61 78 + 39 122 =0.6