Question

Calculate the freezing point of a solution when 3 gram of calcium chloride m equals to 100 gram per mole was dissolved in 100 gram of water assuming cacl2 undergoes complete ionization kf for water equals to 1.86 k kg per mole

Answer 1

Answer: 271.64 K

Explanation:

Given data:

Kf for water = 1.86 kg/mol

Mass of solute (CaCl2) = 3 g

Mass of solvent (water) = 100 g

Molecular mass of CaCl2 = 110.98 g/mol

To find: New freezing point of the solution

The Molality of solution, m  

= (Moles of solute) / (kg of solvent)

= (3/110.98) * (1000/100)

= 0.27 m

Also, given that CaCl2 undergoes complete ionisation, therefore, Vant Hoff Factor “i” will be equal to total no. of dissociated ions i.e.,  

i for CaCl2 = 1+2 = 3

Now,  

The depression in freezing point is given as,

∆Tf = i * Kf * m = 3 * 1.86 * 0.27 = 1.506 K

Thus,  

The new freezing point of the solution = 273.15 – 1.506 = 271.64 K

Answer 2

Answer:

Explanation:

Given Calculate the freezing point of a solution when 3 gram of calcium chloride m equals to 100 gram per mole was dissolved in 100 gram of water assuming cacl2 undergoes complete ionization kf for water equals to 1.86 k kg per mole

Given molecular mass of calcium chloride = 110.98 = 111 gm / mol

So number of moles = 3/111 = 0.03 moles

Now molality = n / W = 0.03 / 100 x 10^-3 = 0.3

Calcium chloride has a van’t hoff factor of 3 = i

So depression in freezing point Δ Tf = i x Kf x molality

                                                        = 3 x 1.86 x 0.3

                                                        = 1.674

So freezing point of solution is – 1.674 degree C