Calculating the molar fraction

42 g of methanol (CH3OH) was dissolved in 111.9 g of water. Calculate the molar fraction of methanol in the resulting solution.

This is a bit confusing, can anybody solve?

First, you need to calculate the amount of substance (n):

n(CH3OH) = m(CH3OH)/M(CH3OH) = (42 g)/(32.04 g/mol) = 1.3 mol
n(H2O) = m(H2O)/M(H2O) = (111.9 g)/(18.02 g/mol) = 6.210 mol
n(total) = n(CH3OH) + n(H2O) = 1.3 mol + 6.210 mol = 7.5 mol

The MOLE (molar is incorrect) fraction of CH3OH is then:

x(CH3OH) = n(CH3OH)/n(total) = (1.3 mol)/(7.5 mol) = 0.17 (or 17 %)

It should not be, let us try it, mol wt of methanol = 32, so moles of methanol = 42/32 = 1.3125
moles of water = 111.9/18 = 6.2167
Total no. of moles = 1.3125 + 6.2167 = 7.5292
molar fraction of methanol = 1.3125/7.5292 = 0.174

There you have it