A solution is prepared by adding 16. g of CH₃OH (molar mass 32 g/mol) to 90. g of H₂O (molar mass 18 g/mol). The mole fraction of CH₃OH in this solution is closest to which of the following ?
The mole fraction of methanol in a mixture of 16. g of methanol and 90. g of water is 0.091.
Given the mass (m) of a compound and its molar mass (M), we can calculate the number of moles (n) using the following expression.
[tex]n = \frac{m}{M}[/tex]
We will use this expression to calculate the moles of methanol and water.
[tex]nCH_3OH = \frac{16. g}{32 g/mol} = 0.50 mol\\nH_2O = \frac{90. g}{18 g/mol} = 5.0 mol[/tex]
The total number of moles is:
[tex]n = 0.50 mol + 5.0 mol = 5.5 mol[/tex]
The mole fraction of methanol is the ratio of the moles of methanol and the total number of moles.
[tex]\chi CH_3OH = \frac{0.50 mol}{5.5 mol} = 0.091[/tex]
The mole fraction of methanol in a mixture of 16. g of methanol and 90. g of water is 0.091.
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