CHEMISTRY HIGH SCHOOL

When 70.4 g of benzamide (C,H,NO) 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. Calculate the van't Hoff factor for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

Answers

Answer 1

Answer:

Answer:

1.60 is the van't Hoff factor for ammonium chloride in X.

Explanation:

$\Delta T_f=iK_f* m$

$Delta T_f=K_f* \frac{\text{Mass of solute}}{\text{Molar mass of solute}* \text{Mass of solvent in Kg}}$ ...(1)

where,

$\Delta T_f$ =Elevation in boiling point =

i = van't Hoff factor

$K_f$ = 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

$K_f$ = Freezing point constant of liquid X= ?

$\Delta T_f=2.7^oC$

Putting all value in a (1):

$2.7^oC=K_f* (70.4 g)/(121 g/mol* 0.850 kg)$

$K_f=3.944 ^oC kg/mol$

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

$K_f=3.944 ^oC kg/mol$

$\Delta T_f=9.9^oC$

Putting all value in a (1):

$9.9^oC=i* 3.944^oC kg/mol* (70.4 g)/(53.5 g/mol* 0.850 kg)$

i = 1.6011 ≈ 1.60

1.60 is the van't Hoff factor for ammonium chloride in X.

Answer 2

Answer:

Final answer:

The van't Hoff factor, which measures ionization, for ammonium chloride in the mysterious liquid X can be calculated to be approximately 1.01. This is calculated by first determining the cryoscopic constant from the observed depression of the freezing point by benzamide (which does not ionize), and then utilizing this value to calculate the theoretical freezing point depression for ammonium chloride (pretending it does not ionize either). Since the observed depression was 9.9℃ and the calculated was 9.8℃, the van't Hoff factor is their quotient, or approximately 1.01.

Explanation:

To solve this problem, we need to understand that the van't Hoff factor (i) is a measure of the extent of ionization in solution. It can be calculated using the formula i = ΔTf observed / ΔTf calculated, where ΔTf observed is the observed freezing point depression and ΔTf calculated is the theoretical freezing point depression if no ionization occurs.

First, we calculate the theoretical freezing point depression for ammonium chloride. We know that this is given by the benzamide that reduces the freezing point of the same amount of liquid X by 2.7℃. Therefore we assume the van't Hoff factor of benzamide is 1 (since it does not ionize) and we get the cryoscopic constant (Kf) of X from ΔTf = Kf * m * i. Substituting into the formula and rearranging gives Kf = ΔTf / (m * i) = 2.7 ℃/(70.4 g/850 g) = 2.7 ℃/0.082824 = 32.6 ℃ kg/mol.

We then use this Kf to calculate the ΔTf calculated for ammonium chloride: ΔTf calculated = Kf * m * i (where we again assume i=1) = 32.6 ℃ kg/mol * (70.4 g/850 g) = 9.8 ℃. Finally we can calculate the van't Hoff factor for ammonium chloride using the original formula: i = ΔTf observed / ΔTf calculated = 9.9 ℃ / 9.8 ℃ = 1.01.

Learn more about Van't Hoff Factor here:

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