Question

A sample of acetic acid solution has a freezing point depression of 2.0 C. What is the e estimated percent by mass of acetic acid in the acetic acid solution? Molar mass of acetic acid C2H3O2=60g/mol.

Answer 1

Freezing point depression = kf*m*i

i= 1 for acetic acid, m= molality

Kf= Freezing point depression constant for acetic acid =3.91 deg.c/molal

m= molality=

2= 3.91*m*1

m= 2/3.91=0.5115

moles of solute/ kg of solvent= 0.5115

Basis : 1 kg of watet (solvent) =1000gm of solvent

Moles of solute acetic acid= 0.5115*1 =0.5115 moles

Molecular weigh of acetic acid =60

Mass of acetic acid = Moles* Molecular weight= 0.5115*60= 30.69 gms

Total mass of solution = mass of water + mass of acetic acid = 1000+30.69 =1030.69 gm

Mass percent of acetic acid = 100*30.69/1030.89=2.97%