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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.

Sagot :

ajeigbeibraheem

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  [tex]=\dfrac{70.4 \ g}{121.14 \ g/mol}[/tex]

= 0.58 mole

The molality = [tex]\dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent }[/tex]

[tex]= \dfrac{0.58 }{0.85 }[/tex]

= 0.6837

Using the formula:

[tex]\mathbf {dT = l \times k_f \times m}[/tex]

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 = [tex]\dfrac{70.4 \ g}{53.491 \ g /mol}[/tex]

= 1.316 mol

The molality = [tex]\dfrac{1.316 \ mol}{0.85 \ kg}[/tex]

= 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

[tex]l = \dfrac{9.9}{3.949 \times 1.5484 \ m}[/tex]

l = 1.62

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