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=(46.85)/(299.4 g/mol)\\\\mol=0.1565`

mol Ethanol, C₂H₅OH(MW=46.07 g/mol)

`\tt mol=(125.5)/(46.07)=2.7241`

mol of solution :

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

• mol fraction Codein :

`\tt (0.1565)/(2.8806)=0.054`

mol fraction Ethanol :

`\tt (2.7241)/(2.8806)=0.946`