Question

Calculate the mole fraction of each solute and solvent in the solution of 46.85 g of codeine, C 18 H 21 NO 3 , in 125.5 g of ethanol, C 2 H 5 OH.

Answer 1

mol fraction Codein : 0.054

mol fraction Ethanol : 0.946

Further explanation

Given

46.85 g of codeine

125.5 g of ethanol

Required

mol fraction

Solution

The mole fraction : the mole ratio of a substance to the mole of solution /mixture  

mol Codeine, C₁₈H₂₁NO₃(MW=299.4 g/mol) :

$\tt mol=\dfrac{46.85}{299.4 g/mol}\\\\mol=0.1565$

mol Ethanol, C₂H₅OH(MW=46.07 g/mol)

$\tt mol=\dfrac{125.5}{46.07}=2.7241$

mol of solution :

$\tt mol~codein+mol~ethanol=0.1565+2.7241=2.8806$

• mol fraction Codein :

$\tt \dfrac{0.1565}{2.8806}=0.054$

mol fraction Ethanol :

$\tt \dfrac{2.7241}{2.8806}=0.946$