A solution is prepared by mixing 4.00 grams of benzene (molar mass = 78.1 g/mol), 4.00 grams of CCl₄ (molar mass = 153.8 g/mol), and 4.00 grams of C S₂ (molar mass = 76.1 g/mol). The mole fraction of benzene in this solution is:

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of each component in the solution:

Moles of benzene = 4.00 g / 78.1 g/mol = 0.0512 mol
Moles of CCl4 = 4.00 g / 153.8 g/mol = 0.026 mol
Moles of CS2 = 4.00 g / 76.1 g/mol = 0.0525 mol

Next, we calculate the total moles of solute in the solution:

Total moles of solute = 0.0512 + 0.026 + 0.0525 = 0.1297 mol

Finally, we calculate the mole fraction of benzene:

Mole fraction of benzene = moles of benzene / total moles of solute = 0.0512 / 0.1297 ≈ 0.395

Therefore, the answer is D, with a mole fraction of benzene of approximately 0.395.

A Solution Is Prepared by Mixing 4