Calculate the molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of

1.	Calculate the molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene.
Answer» Molecular mass of ethanoic acid=60 Molecular mass of benzene=78 Number of moles of ethanoic acid= 2.5/60=0.041Number of moles of benzene= 75/78=0.96153Mole fraction= 0.041/(0.041+0.96153)=0.04 Molality = Number of moles of solute/weight of the solvent (in kg)= 0.041/0.075=0.5466

Calculate the molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene.

Answer»

Molecular mass of ethanoic acid=60

Molecular mass of benzene=78

Number of moles of ethanoic acid= 2.5/60=0.041Number of moles of benzene= 75/78=0.96153Mole fraction= 0.041/(0.041+0.96153)=0.04

Molality = Number of moles of solute/weight of the solvent (in kg)= 0.041/0.075=0.5466