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it takes 3.11 ml of water to fully dissolve 4.54 grams of urea to form …

Question

it takes 3.11 ml of water to fully dissolve 4.54 grams of urea to form a saturated solution at a temperature of 20.13 °c. what is the change in free energy (in units of j/mol) of this process? report your answer to the nearest whole number. note: this example is based on a small range of actual data. your experimental numbers may or may not fall within this range. the purpose of this question is to be an example and to give you practice in performing the calculation.

Explanation:

Step 1: Find the moles of urea

The molar mass of urea ($\ce{CO(NH_{2})_{2}}$) is calculated as follows:

  • Carbon (C): 12.01 g/mol
  • Oxygen (O): 16.00 g/mol
  • Nitrogen (N): 14.01 g/mol (2 atoms)
  • Hydrogen (H): 1.008 g/mol (4 atoms)

Molar mass of urea, $M = 12.01 + 16.00 + 2\times(14.01) + 4\times(1.008) = 60.06\ \text{g/mol}$

Moles of urea, $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.54\ \text{g}}{60.06\ \text{g/mol}} \approx 0.0756\ \text{mol}$

Step 2: Find the volume of the solution (approximate as volume of water)

The volume of water is $3.11\ \text{mL} = 3.11\times10^{-3}\ \text{L}$

Step 3: Calculate the molarity (concentration) of the saturated solution

Molarity, $c = \frac{n}{V} = \frac{0.0756\ \text{mol}}{3.11\times10^{-3}\ \text{L}} \approx 24.31\ \text{M}$ (Wait, this seems very high. Wait, maybe we need to consider the activity or use the concept of solubility. Wait, actually, for a saturated solution, the chemical potential is equal to the chemical potential of the solid, so the change in free energy, $\Delta G = 0$? Wait, no, maybe I made a mistake. Wait, no, when a solution is saturated, the process of dissolving is at equilibrium, so $\Delta G = 0$? But that can't be. Wait, maybe the question is about the dissolution process, but if it's saturated, the equilibrium constant is related to the solubility. Wait, maybe I misread the question. Wait, the question says "the change in free energy of this process" – the process of dissolving urea to form a saturated solution. At equilibrium (saturated solution), $\Delta G = 0$. But that seems too easy. Wait, maybe there's a miscalculation in molarity. Wait, 4.54 grams of urea in 3.11 mL of water. Let's recalculate molarity:

Moles of urea: $4.54\ \text{g} / 60.06\ \text{g/mol} \approx 0.0756\ \text{mol}$

Volume of solution: approximately 3.11 mL (since the volume of solute is negligible compared to solvent). So volume in liters: $3.11\times10^{-3}\ \text{L}$

Molarity: $0.0756\ \text{mol} / 0.00311\ \text{L} \approx 24.3\ \text{M}$. But the solubility of urea in water at 20°C is about 1080 g/L, which is $1080\ \text{g/L} / 60.06\ \text{g/mol} \approx 18\ \text{M}$. Wait, so 24.3 M is higher than the actual solubility, which means maybe the data is hypothetical. But regardless, for a saturated solution at equilibrium, $\Delta G = 0$ because the system is at equilibrium (the rate of dissolution equals the rate of crystallization). So the change in free energy should be 0 J/mol.

Wait, maybe the question is not about equilibrium? But it says "to form a saturated solution", which is an equilibrium state. So $\Delta G = 0$.

Answer:

0