A 1.00 mol sample of CH3OH (l) was added to an infinitely large sample of CH3OH-

Question

A 1.00 mol sample of CH3OH (l) was added to an infinitely large sample of CH3OH-H20 solution with XCH3OH=0.0900. For this solution, VH2O = 18.1 ml/mol and VCH3OH = 52.2 ml/mol.

1) What was the change in volume for the process?

**I believe there to be no change as an infinitely large sample would be infinite and x+infinity will always yield infinity. ** Correct me if I'm wrong.

2) Determine the volume of pure liquids needed to prepare 100mL of a solution with XCH3OH = 0.090. The densities of the pure liquids are (wrong for methanol, but I digress) 0.7893 g/mL for methanol and 0.9970 for water.

Explanation / Answer

Yeah you are correct about the first answer

Vol of methanol = 0.09 x 100 = 9 mL

Mass = 9 x 0.7893 = 7.1037 g

Volume of water = 91 mL

Mass = 91 x 0.9970 = 90.727 g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.