Question

29. calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride

Answer 1

Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 - 30)g = 70 g

Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol - 1

= 78 g mol - 1

∴ Number of moles of C6H6  =30/78 mol

= 0.3846 mol

Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5

= 154 g mol - 1

∴ Number of moles of CCl4 = 70/154 mol

= 0.4545 mol

Thus, the mole fraction of C6H6 is given as:

NO. OF MOLES OF C6H6 / TOTAL NO. OF MOLES OF BOTH COMPONDS

= 0.3846 / (0.3846 +0.4545)

= 0.458

Answer 2

Hey !!

Let the mass of solution be 100 g

Mass of Benzene = 30 g

Mass of carbon tetrachloride = 100 g - 30 g = 770 g

Number of moles of benzene = Mass / Molar mass

                     = 30 g / 78 g mol⁻¹

                     = 0.385 mol

Number of moles of carbon tetrachloride = Mass / Molar mass

                   = 70 g / 154 g mol⁻¹

                   = 0.455 mol

Mole fraction of benzene = Mass of benzene / Moles of benzene + Moles of tetrachloride

                 = 0.385 mol / 0.840 mol

FINAL RESULT  :- 0.458

Hope it helps you !!