Question

Assuming that CaCl2 (MW = 111.0 g/mol) dissociates 100% in solution, what mass of CaCl2 is required to lower the freezing point of 500.0 g water to -5.0 oC? [Kf = 1.86 oC Kg /mol]

Answer 1

We know

∆T=Kf×m

Where ∆T=change in temperature=-5.0°C

Kf=freezing point depression constant=1.86°CKg/mol

m=molality

5.0=1.86×m

=>m=2.688mol/kg

Mass of solvent=500g=0.5Kg

We know m=no. of moles/mass of solvent,n=no. of moles

2.688=n/0.5

=>n=1.344

We know,

n=mass/Molar mass

Molar mass of Cacl2=110.0g/mol

mass of CaCl2=n×Molar mass

Mass of CaCl2=1.344×110.0g

=147.84g

Hence,147.84g of CaCl2