# Chapter 13 - Properties of Solutions - Exercises: 13.49a

$X_{CH_3OH}=0.227$

#### Work Step by Step

First, we need to find moles of methanol and moles of acetonitrile. Also, from the periodic table, molar mass of methanol = 32.04 g/mol. Molar mass of acetonitrile = 41.05 g/mol $\frac{22.5\;mL\;methanol}{} (\frac{0.791\;g}{1 \;mL})(\frac{1\;mol}{32.04\;g})=0.5555\;mol\;methanol$ $\frac{98.7\;mL\;acetonitrile}{}(\frac{0.786\;g}{1\;mL})(\frac{1\;mol}{41.05\;g})=1.890\;mol\;acetonitrile$ Now, we will find the mole fraction of methanol. That is, the number of moles of methanol divided by total moles. $X_{methanol}=\frac{0.5555\;mol\;methanol}{0.5555\;mol\;methanol+1.890\;mol\;acetonitrile}=0.227$

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