Problem 50
Question
A A sample of uranium fluoride is found to effuse at the rate of \(17.7 \mathrm{mg} / \mathrm{h}\). Under comparable conditions, gaseous \(\mathrm{I}_{2}\) effuses at the rate of \(15.0 \mathrm{mg} / \mathrm{h} .\) What is the molar mass of the uranium fluoride? (Hint: Rates must be converted to units of moles per time.
Step-by-Step Solution
Verified Answer
The molar mass of uranium fluoride is approximately 182.42 g/mol.
1Step 1: Understand Graham's Law of Effusion
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, if two gases, A and B, have rates of effusion (Rate_A and Rate_B) and molar masses (M_A and M_B), the following relationship holds: \( \frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}} \).
2Step 2: Set Up the Equation Using Given Rates
We are given the rates of effusion for uranium fluoride (Rate_UF) and iodine \((\text{Rate}_{I_2})\). These are 17.7 mg/h and 15.0 mg/h, respectively. Substitute these into Graham's Law: \( \frac{17.7}{15.0} = \sqrt{\frac{M_{I_2}}{M_{UF}}} \).
3Step 3: Calculate the Molar Mass of Iodine
To proceed further, first determine the molar mass of iodine \((\text{I}_2)\). Since iodine has an atomic mass of approximately 127 u, the molar mass of \(\text{I}_2\) is \({2 \times 127 = 254 \, \text{g/mol}}\).
4Step 4: Solve for Molar Mass of Uranium Fluoride
Insert the molar mass of iodine into the equation: \( \frac{17.7}{15.0} = \sqrt{\frac{254}{M_{UF}}} \). Square both sides and solve for \(M_{UF}\): \( \left(\frac{17.7}{15.0}\right)^2 = \frac{254}{M_{UF}} \). Simplifying yields \(M_{UF} = \frac{254}{\left(\frac{17.7}{15.0}\right)^2}\).
5Step 5: Calculate Numerically
Perform the calculations: \( \left(\frac{17.7}{15.0}\right)^2 \approx 1.3921 \). Thus, \( M_{UF} \approx \frac{254}{1.3921} \approx 182.42 \, \text{g/mol} \).
Key Concepts
Molar Mass CalculationRate of EffusionEducational Chemistry Problem
Molar Mass Calculation
Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a given substance. It is usually expressed in grams per mole (g/mol). Calculating molar mass is crucial because it allows us to convert between the mass of a substance and the amount of substance in moles, which is essential for performing stoichiometric calculations in chemical reactions. To calculate the molar mass of a compound:
- Sum up the atomic masses of each element in the compound.
- Multiply the atomic masses by the number of each type of atom present in the formula.
Rate of Effusion
Effusion is the process by which gas molecules escape through a tiny hole into a vacuum. Graham's Law of Effusion provides a way to compare the rates of effusion between two gases. It states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.In formula form, Graham's Law is represented as:\[\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}},\]where \(\text{Rate}_A\) and \(\text{Rate}_B\) are the rates of effusion for gases A and B, and \(M_A\) and \(M_B\) are their respective molar masses.Graham's Law allows us to predict how fast a gas will effuse compared to another. For instance, in the exercise given, uranium fluoride and iodine are compared. By rearranging the law's formula, you can solve for unknown molar masses or rates once the others are known.
Educational Chemistry Problem
Solving chemistry problems effectively requires a good grasp of fundamental principles and the ability to apply them. The exercise involving uranium fluoride and iodine is a perfect example. This type of educational chemistry problem challenges students to apply their understanding of Graham's Law of Effusion, alongside the skill of molar mass calculation.
To solve the uranium fluoride problem, students need to:
- Identify the relationship between two gases given by Graham's Law.
- Convert provided data, like effusion rates, to the necessary units if required.
- Perform calculations accurately by carefully substituting values into the equations.
Other exercises in this chapter
Problem 48
Argon gas is 10 times denser than helium gas at the same temperature and pressure. Which gas is predicted to effuse faster? How much faster?
View solution Problem 49
A gas whose molar mass you wish to know effuses through an opening at a rate one third as fast as that of helium gas. What is the molar mass of the unknown gas?
View solution Problem 51
Under which set of conditions will \(\mathrm{CO}_{2}\) deviate most from ideal gas behavior? (a) \(1 \mathrm{atm}, 0^{\circ} \mathrm{C}\) (b) 0.1 atm, \(100^{\c
View solution Problem 52
Under which set of conditions will \(\mathrm{Cl}_{2}\) deviate least from ideal gas behavior? (a) \(1 \mathrm{atm}, 0^{\circ} \mathrm{C}\) (b) 0.1 atm, \(100^{\
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