Question

A 40.0 gram-sample of methanol, CH4O is mixed with 60.0 grams of ethanol, C2H60. What is the mole fraction of the methanol?

Answer 1

Solution.

$m(CH_4O)=40.0g;$

$m(C_2H_6O)=60.0g;$

$M(CH_4O)=32 g/mol;$

$M(C_2H_6O)=46 g/mol;$

$\nu(CH_4O)=\dfrac{40.0}{32}=1.25mol;$

$\nu(C_2H_6O)=\dfrac{60.0}{46}=1.3mol;$

$X_i=\dfrac{1.25}{1.25+1.3}=0.49;$

Answer: $X_i=0.49.$