Question

At \(27^{\circ} \mathrm{C}, 500 \mathrm{~mL}\) of helium diffuses in 30 minutes. What is the time (in hours) taken for \(1000 \mathrm{~mL}\) of \(\mathrm{SO}_{2}\) to diffuse under same experimental conditions? (a) 240 (b) 340 (c) 200 (d) 440

Short answer

The time taken for 1000 mL of \(\mathrm{SO}_{2}\) to diffuse is 4 hours.

Step-by-step solution

Establish known values and formula

We know that helium diffuses 500 mL in 30 minutes. We are given that 1000 mL of \(\text{SO}_2\) needs to diffuse under the same conditions. We can use Graham's law of effusion to relate the diffusion rates of two gases: \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \] where \(r_1\) and \(r_2\) are the rates of effusion, and \(M_1\) and \(M_2\) are the molar masses of helium and \(\text{SO}_2\), respectively.

Calculate the molar masses

Helium has a molar mass of approximately 4 g/mol, while \(\text{SO}_2\) has a molar mass of \(32 + (2 \times 16) = 64 \text{ g/mol}\).

Determine rates of diffusion

Using Graham's law, first find the rate of diffusion for helium, \(r_1\), as \(\frac{500 \text{ mL}}{30 \text{ mins}} = 16.67 \text{ mL/min}\). Let the rate of diffusion of \(\text{SO}_2\) be \(r_2\). Use the formula: \[ r_2 = r_1 \times \sqrt{\frac{M_1}{M_2}} = 16.67 \times \sqrt{\frac{4}{64}} = 16.67 \times 0.25 = 4.1675 \text{ mL/min} \]

Calculate time for 1000 mL of SO2

Using the rate \(r_2 = 4.1675 \text{ mL/min}\), calculate the time required to diffuse 1000 mL of \(\text{SO}_2\):\[ \text{Time} = \frac{1000 \text{ mL}}{4.1675 \text{ mL/min}} \approx 240 \text{ minutes}\]. Convert this time into hours: \(\frac{240}{60} = 4 \text{ hours}\).

Key concepts

Molar Mass Calculation

Molar mass is a critical component when dealing with chemical calculations, especially in gases. It determines the amount of substance present in a given volume or weight. The molar mass is expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you must add up the atomic masses of each element present in the compound according to the periodic table.

The atomic mass of helium is approximately 4 g/mol. Helium is a single element gas, which simplifies its molar mass calculation. On the other hand, sulfur dioxide ( SO_2 ) has a more complex structure consisting of one sulfur atom and two oxygen atoms. The atomic mass of sulfur is about 32 g/mol and that of oxygen about 16 g/mol.

• To find the molar mass of SO_2 , add the mass of one sulfur atom and two oxygen atoms: 32 + (2 x 16) = 64 g/mol.

This calculation of molar mass is essential for applying Graham's Law, where we compare the molar masses of two gases to understand their diffusion rates.

Diffusion Rate Calculation

Diffusion rate refers to how fast gas molecules spread throughout a space. Graham's Law of Diffusion is used to compare the rates at which two different gases will diffuse.

According to Graham's Law, the diffusion rate (r) is inversely proportional to the square root of its molar mass (M). This principle can be expressed with the formula: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \). Here, \( r_1 \) and \( r_2 \) represent the diffusion rates of two gases, while \( M_1 \) and \( M_2 \) are their respective molar masses.

• For helium, which diffuses at a known rate of 500 mL in 30 minutes, the diffusion rate is calculated as 16.67 mL/minutes.
• The diffusion rate of SO_2 is determined using Graham's Law, based on knowing helium's diffusion rate and molar masses. For this exercise, SO_2 diffuses at approximately 4.1675 mL/minute.

This calculation demonstrates how diffusion rates vary inversely with the mass of gas molecules, making it a fundamental part of predicting gas behaviors in different conditions.

Stoichiometry in Gases

Stoichiometry involves using balanced chemical equations to calculate the quantities of reactants and products involved in chemical reactions. While it often is associated with reactions, it also plays a crucial role in understanding the behavior of gases.

In the context of gas diffusion, stoichiometry helps in determining how much of one gas will diffuse in relation to another under the same conditions. By understanding the diffusion rates and molar masses, we can apply stoichiometry to compare gas quantities.

• In our example, helium diffuses at 16.67 mL/minute, while SO_2 diffuses slower at about 4.1675 mL/minute.
• Knowing these rates allows calculation of how much time it will take for a certain volume of gas to diffuse, emphasizing stoichiometry’s role in such calculations.

By applying stoichiometric principles to gas diffusion, students can predict and understand the time required for specific volumes of gases to diffuse. This solidifies the relationship between chemical quantities and their physical manifestations, especially in gases.