Respuesta :
Answer:
1.60 is the van't Hoff factor for ammonium chloride in X.
Explanation:
[tex]\Delta T_f=iK_f\times m[/tex]
[tex]Delta T_f=K_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]...(1)
where,
[tex]\Delta T_f[/tex] =Elevation in boiling point =
i = van't Hoff factor
[tex]K_f[/tex] = Freezing point constant
m = molality
1) When 70.4 g of benzamide are dissolved in 850. g of a certain mystery liquid X.
Mass of benzamide = 70.4 g
Molar mass of benzamide = 121 g/mol
i = 1 (organic molecule)
Mass of liquid X = 850 g = 0.850 kg
[tex]K_f[/tex] = Freezing point constant of liquid X= ?
[tex]\Delta T_f=2.7^oC[/tex]
Putting all value in a (1):
[tex]2.7^oC=K_f\times \frac{70.4 g}{121 g/mol\times 0.850 kg}[/tex]
[tex]K_f=3.944 ^oC kg/mol [/tex]
2) When 70.4 g of ammonium chloride are dissolved in 850. g of a certain mystery liquid X.
Mass of ammonium chloride= 70.4 g
Molar mass of ammonium chloride = 53.5 g/mol
i = ? (ionic molecule)
Mass of liquid X = 850 g = 0.850 kg
[tex]K_f=3.944 ^oC kg/mol[/tex]
[tex]\Delta T_f=9.9^oC[/tex]
Putting all value in a (1):
[tex]9.9^oC=i\times 3.944^oC kg/mol\times \frac{70.4 g}{53.5 g/mol\times 0.850 kg}[/tex]
i = 1.6011 ≈ 1.60
1.60 is the van't Hoff factor for ammonium chloride in X.
