Mole fraction: A solution contains `127g` of merthanot `(CH_(3) OH)` in `108g` of water `(H_(2) O)`. What are the mole fractions of `CH_(3) OH` and `H_(2) O`?
Strategy: Convert the masses of both components to their moles, and them apply the definiton fo mole fraction.

Explanation

Step 1: Calculate moles.
Number of moles `(n) = ("Mass in grams")/("Molar mass")`
`.^(n)CH_(3) OH = (128g)/(32.0g mol^(-1)) = 4.00 "mol" CH_(3) OH`
`.^(n)H_(2)O = (108g)/(18.0 g mol^(-1)) = 6.00 "mol" H_(2) O`
Step 2 : Apply the definition of mole fraction
`chi CH_(3) OH = (.^(n)CH_(3) OH)/(.^(n)CH_(3) OH + .^(n)H_(2)O) = (4.00 mol)/((4.00+ 6.00) "mol") = 0.400`
`chi H_(2) O = (.^(n)H_(2) O)/(.^(n)CH_(3) OH +. ^(n)H_(2) O) = (6.00 mol)/((4.00 + 6.00) "mol") = 0.600`