When 48.7g of benzamide C7H7NO are dissolved in 1300.g of a certain mystery liquid X, the freezing point of the solution is 2.3°C lower than the freezing point of pure X. On the other hand, when 48.7g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 8.7°C lower than the freezing point of pure X.Calculate the van’t Hoff factor for sodium chloride inX. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits. Please dont round untill the end.
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The Correct Answer and Explanation is:
To calculate the van’t Hoff factor (i) for sodium chloride (NaCl) in solvent X, we compare the freezing point depression it causes with that caused by benzamide, a non-electrolyte (i = 1).
Step-by-Step Calculation
1. Freezing Point Depression Equation:
ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m
Where:
- ΔTf\Delta T_f = freezing point depression (°C)
- ii = van’t Hoff factor
- KfK_f = cryoscopic constant (same for both solutions)
- mm = molality of solute
We will use benzamide to determine KfK_f because its van’t Hoff factor is known (i = 1).
For Benzamide:
Given:
- Mass of benzamide = 48.7 g
- Molar mass of benzamide (C₇H₇NO) ≈ 121.14 g/mol
- Mass of solvent = 1300 g = 1.300 kg
- ΔTf=2.3∘C\Delta T_f = 2.3^\circ C
Molality (m): mol benzamide=48.7121.14≈0.402mol\text{mol benzamide} = \frac{48.7}{121.14} \approx 0.402 mol m=0.4021.300≈0.309 mol/kgm = \frac{0.402}{1.300} \approx 0.309 \, \text{mol/kg}
Now use: ΔTf=i⋅Kf⋅m=1⋅Kf⋅0.309=2.3\Delta T_f = i \cdot K_f \cdot m = 1 \cdot K_f \cdot 0.309 = 2.3 K_f = \frac{2.3}{0.309} \approx 7.44 \, ^\circ C \cdot \text{kg/mol}
For NaCl:
Given:
- Mass NaCl = 48.7 g
- Molar mass NaCl ≈ 58.44 g/mol
- ΔTf=8.7∘C\Delta T_f = 8.7^\circ C
Molality: mol NaCl=48.758.44≈0.833mol\text{mol NaCl} = \frac{48.7}{58.44} \approx 0.833 mol m=0.8331.300≈0.641 mol/kgm = \frac{0.833}{1.300} \approx 0.641 \, \text{mol/kg}
Now solve for ii: ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m 8.7=i⋅7.44⋅0.6418.7 = i \cdot 7.44 \cdot 0.641 i=8.77.44⋅0.641≈8.74.77≈1.823i = \frac{8.7}{7.44 \cdot 0.641} \approx \frac{8.7}{4.77} \approx 1.823
Final Answer:
i=1.8\boxed{i = 1.8}
Explanation
The van’t Hoff factor (i) quantifies how many particles a solute produces in solution, and is essential for calculating colligative properties like freezing point depression. For non-electrolytes like benzamide, i = 1 because they do not dissociate in solution. Ionic compounds, such as NaCl, dissociate into ions—in this case, ideally into Na⁺ and Cl⁻, suggesting a theoretical i = 2. However, in real solutions, ion pairing and incomplete dissociation reduce the effective van’t Hoff factor.
In this problem, both benzamide and NaCl are dissolved in the same mass of solvent X. We first calculate the molality of each solute. Using the known freezing point depression for benzamide and the assumption that it behaves ideally (i = 1), we determine the freezing point depression constant KfK_f for the solvent.
Then, using the same KfK_f, we calculate the molality of NaCl and plug these into the freezing point depression formula to solve for the van’t Hoff factor. The result, i≈1.8i \approx 1.8, shows that NaCl mostly but not completely dissociates, which is typical in non-ideal solutions. This effective van’t Hoff factor reflects real-world conditions like ion-ion interactions or solvent effects that reduce dissociation.
This approach allows us to experimentally determine how many particles are influencing the colligative property, providing insight into the ionic behavior of solutes in solution.
