Question

What are the mole fractions of methanol (CH3OH) and water (H2O) in the solution that contains 128g of CH3OH 108g of water?

Answer 1

Answer:

XMeOH = 0.40....

Explanation:

XMeOH = $\frac{128.g}{32.04.g .mol^{-1} }\\\frac{128.g}{32.04 .g . mol^{-1} } +\frac{108.g}{18.01 .g .mol^{-1} }$

Now we know that in a binary solution, the sum of the mole fractions will be one....but let us belabour the point........

Xwater = $\frac{108.g}{180.1 .g .mol^{-1} } \\\frac{128.g}{32.04 .g .mol^{-1} } + \frac{108.g}{18001 .g .mol^{-1} }$

and

XMeOH Xwater = 1 required..