Question

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

Answer 1

Answer:

$\underline{\Huge{\texttt{0.5466}}}$

Explanation:

Molecular mass of ethanoic acid=60

Molecular mass of benzene=78

Number of moles of ethanoic acid= 2.5/60=0.041

Number of moles of benzene= 75/78=0.96153

Mole 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