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

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

Step-by-step explanation:

`\Delta T_f=i* K_f* 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}* \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* K_f * (63.4 g)/(71 g/mol* 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* 6.19^oC* (63.4 g)/(162.5 g/mol* 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.