Question

When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2.7 C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH CI) are dissolved in the same mass of X, the freezing point of the solution is 9.9 °C lower than the freezing point of pure X.

Required:
Calculate the van't Hoff factor for ammonium chloride in X.

Answer 1

1.62

Step-by-step explanation:

From the given information:

number of moles of benzamide
`=(70.4 \ g)/(121.14 \ g/mol)`

= 0.58 mole

The molality =
`(mass \ of \ solute (i.e. \ benzamide ))/(mass \ of \ solvent )`

`= (0.58 )/(0.85 )`

= 0.6837

Using the formula:

`\mathbf {dT = l * k_f * m}`

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf × 0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl =
`(70.4 \ g)/(53.491 \ g /mol)`

= 1.316 mol

The molality =
`(1.316 \ mol)/(0.85 \ kg)`

= 1.5484

Thus;

the above kf value is used in determining the Van't Hoff factor for NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

`l = (9.9)/(3.949 * 1.5484 \ m)`

l = 1.62