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
