1.

A solution has been prepared by dissolved 60 g of methyl alcohol in 120 g of water. What is the mole fraction of methyl alcohol and water ?

Answer»


SOLUTION :MOLES of `CH_(3)OH=("Mass of "CH_(3)OH)/("MOLAR mass")=((60 G)/("32 g mol"^(-1)))=1.875` mol
Moles of water `= ("Mass of water")/("Molar mass")=((120g)/("18 g mol"^(-1)))=6.667` mol
Mole fraction of `CH_(3)OH=(("1.875 mol"))/(("1.875 mol +6.667 mol"))=(("1.875 mol"))/(("8.542 mol"))=0.220`
Mole fraction water `= 1 - 0.220 = 0.780`.


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