Question

When 63.4 g of glycine (C2HNO2 are dissolved in 700. g of a certain mystery liquid X, the freezing point of the solution is 7.9 °C lower than the freezing point of pure X. On the other hand, when 63.4 g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 13.3 °C lower than the freezing point of pure X Calculate the van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits. x 10

Answer 1

Answer:

3.8 is the van't Hoff factor for iron(III) chloride in X.

Explanation:

$\Delta T_f=i\times K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality =$\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

i = van't Hoff factor

we have :

Mass of glycine = 63.4 g

Molar mass of glycine = 71 g/mol

Mass of solvent X = 700. g = 0.7 kg

1 g = 0.001 kg

$K_f$ of solvent X= ?

i = 1  (non electrolyte)

Depression in freezing point= $7.9^oC$

$7.9^oC=1\times K_f \times \frac{63.4 g}{71 g/mol\times 0.7 kg}$

$K_f=6.19 ^oC/m$

When iron(III) chloride is dissolved in 0.7 kg of solvent X

Mass of  iron(III) chloride = 63.4 g

Molar mass of  iron(III) chloride= 162.5 g/mol

Mass of solvent X = 700. g = 0.7 kg

1 g = 0.001 kg

$K_f$ of solvent X= $6.19 ^oC/m$

i = ?

Depression in freezing point:$13.3^oC$

$13.3^oC=i\times 6.19^oC\times \frac{63.4 g}{162.5 g/mol\times 0.7 kg}$

Solving for i:

i = 3.85 ≈ 3.8

3.8 is the van't Hoff factor for iron(III) chloride in X.