Graham's Law and Dalton's Law Worksheet

Graham’s Law and Dalton’s Law Worksheet

Question 1: Does NO diffuse faster or slower than CH₄? Why?

Solution:

Graham’s Law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass:

$\text{Rate} \propto \frac{1}{\sqrt{M}}$

To determine whether NO diffuses faster or slower than CH₄, we compare their molar masses:

Molar mass of NO (Nitric oxide):

$M = (14.01 , \text{g/mol for N}) + (16.00 , \text{g/mol for O}) = 30.01 , \text{g/mol}$

Molar mass of CH₄ (Methane):

$M = (12.01 , \text{g/mol for C}) + (4 \times 1.008 , \text{g/mol for H}) = 16.04 , \text{g/mol}$

Since CH₄ has a smaller molar mass than NO, it diffuses faster.

Answer: CH₄ diffuses faster than NO because it has a lower molar mass, and according to Graham’s Law, the lighter gas diffuses more rapidly.

Question 2: Average speed of CO molecules under the same conditions as CO₂.

Solution:

The root mean square speed ( $v_{\text{rms}}$ ) is proportional to the inverse of the square root of the molar mass:

$\frac{v_{\text{CO}}}{v_{\text{CO}2}} = \sqrt{\frac{M{\text{CO}2}}{M{\text{CO}}}}$

Given:

Average speed of CO₂: $v_{\text{CO}_2} = 25.0 , \text{m/s}$
Molar mass of CO₂:
$M_{\text{CO}_2} = (12.01) + (2 \times 16.00) = 44.01 , \text{g/mol}$
Molar mass of CO:
$M_{\text{CO}} = (12.01) + (16.00) = 28.01 , \text{g/mol}$

Substituting into the equation:

$v_{\text{CO}} = v_{\text{CO}2} \times \sqrt{\frac{M{\text{CO}2}}{M{\text{CO}}}} = 25.0 , \text{m/s} \times \sqrt{\frac{44.01}{28.01}}$

$v_{\text{CO}} = 25.0 , \text{m/s} \times \sqrt{1.571}$

$v_{\text{CO}} = 25.0 , \text{m/s} \times 1.253 = 31.3 , \text{m/s}$

Answer: The average speed of CO molecules is 31.3 m/s.

Question 3: Find the molar mass of the unknown gas.

Solution:

From Graham’s Law:

$\frac{\text{Rate}{\text{unknown}}}{\text{Rate}{\text{argon}}} = \sqrt{\frac{M_{\text{argon}}}{M_{\text{unknown}}}}$

The rate is inversely proportional to the time taken:

$\frac{\text{Rate}{\text{unknown}}}{\text{Rate}{\text{argon}}} = \frac{t_{\text{argon}}}{t_{\text{unknown}}}$

Given:

$t_{\text{argon}} = 28.0 , \text{s}$
$t_{\text{unknown}} = 45.0 , \text{s}$
Molar mass of argon ( $M_{\text{argon}}$ ) = 39.95 g/mol

Substituting:

$\frac{28.0}{45.0} = \sqrt{\frac{39.95}{M_{\text{unknown}}}}$

Simplify:

$0.622 = \sqrt{\frac{39.95}{M_{\text{unknown}}}}$

Square both sides:

$0.387 = \frac{39.95}{M_{\text{unknown}}}$

Solve for $M_{\text{unknown}}$ :

$M_{\text{unknown}} = \frac{39.95}{0.387} = 103.2 , \text{g/mol}$

Answer: The molar mass of the unknown gas is 103.2 g/mol.

Question 4: Which odor is smelled first?

Solution:

The substance with the lowest molar mass diffuses the fastest. Calculate the molar masses:

Hydrogen sulfide ( $H_2S$ ):

$M = (2 \times 1.008) + (32.07) = 34.09 , \text{g/mol}$

Methyl salicylate ( $C_8H_8O_3$ ):

$M = (8 \times 12.01) + (8 \times 1.008) + (3 \times 16.00) = 152.16 , \text{g/mol}$

Benzaldehyde ( $C_7H_6O$ ):

$M = (7 \times 12.01) + (6 \times 1.008) + (16.00) = 106.12 , \text{g/mol}$

The rate of diffusion is inversely proportional to the square root of the molar mass. The gas with the smallest molar mass diffuses fastest. Ranking by molar mass:

$H_2S$ : 34.09 g/mol (fastest)
$C_7H_6O$ : 106.12 g/mol
$C_8H_8O_3$ : 152.16 g/mol (slowest)

Answer: The rotten egg odor of $H_2S$ will be smelled first because it has the lowest molar mass and diffuses the fastest.

Question 5: Determine the rate of effusion of the densest gas known, UF₆, to that of the lightest gas, H₂.

Solution:

Using Graham’s Law:

$\frac{\text{Rate}_{\text{UF}6}}{\text{Rate}{\text{H}2}} = \sqrt{\frac{M{\text{H}2}}{M{\text{UF}_6}}}$

Step 1: Calculate the molar masses.

Molar mass of H₂: $M_{\text{H}_2} = 2 \times 1.008 = 2.016 , \text{g/mol}$
Molar mass of UF₆:
$M_{\text{UF}_6} = (238.03 , \text{g/mol for U}) + (6 \times 18.998 , \text{g/mol for F}) = 352.03 , \text{g/mol}$

Step 2: Substitute into Graham’s Law.

$\frac{\text{Rate}_{\text{UF}6}}{\text{Rate}{\text{H}_2}} = \sqrt{\frac{2.016}{352.03}}$

$\frac{\text{Rate}_{\text{UF}6}}{\text{Rate}{\text{H}_2}} = \sqrt{0.00573}$

$\frac{\text{Rate}_{\text{UF}6}}{\text{Rate}{\text{H}_2}} = 0.0757$

Step 3: Interpret the result.
The rate of effusion of UF₆ is 0.0757 times the rate of effusion of H₂, meaning UF₆ effuses much slower.

Answer: The rate of effusion of UF₆ is 0.0757 times that of H₂.

Question 6: An unknown gas diffuses 1.62 times slower than does oxygen gas. Make a reasonable prediction as to what the unknown gas is.

Solution:

Using Graham’s Law:

$\frac{\text{Rate}{\text{unknown}}}{\text{Rate}{\text{O}2}} = \sqrt{\frac{M{\text{O}2}}{M{\text{unknown}}}}$

Rearranging to find the molar mass of the unknown gas:

$M_{\text{unknown}} = M_{\text{O}2} \times \left(\frac{\text{Rate}{\text{O}2}}{\text{Rate}{\text{unknown}}}\right)^2$

Step 1: Calculate the molar mass of O₂.

$M_{\text{O}_2} = 2 \times 16.00 = 32.00 , \text{g/mol}$

Step 2: Substitute values into the equation.
Given that the unknown gas diffuses 1.62 times slower, the rate ratio is

$\frac{\text{Rate}_{\text{O}2}}{\text{Rate}{\text{unknown}}} = 1.62$

$M_{\text{unknown}} = 32.00 \times (1.62)^2$
$M_{\text{unknown}} = 32.00 \times 2.6244$

$M_{\text{unknown}} = 83.98 , \text{g/mol}$

Step 3: Identify the gas.
A gas with a molar mass close to 84 g/mol is likely krypton (Kr), which has a molar mass of 83.80 g/mol.

Answer: The unknown gas is likely krypton (Kr).

Question 7: A container with two gases, helium and argon, is 30.0% by volume helium. Calculate the partial pressure of helium and argon if the total pressure inside the container is 4.00 atm.

Solution:

Dalton’s Law of Partial Pressures states:

$P_{\text{total}} = P_{\text{He}} + P_{\text{Ar}}$

The partial pressure of a gas is proportional to its mole (or volume) fraction:

$P_{\text{He}} = \text{Volume fraction of He} \times P_{\text{total}}$

$P_{\text{Ar}} = \text{Volume fraction of Ar} \times P_{\text{total}}$

Step 1: Determine the volume fractions.

Volume fraction of helium: $30.0% = 0.300$
Volume fraction of argon: $100% - 30.0% = 70.0% = 0.700$

Step 2: Calculate the partial pressures.

Partial pressure of helium:
$P_{\text{He}} = 0.300 \times 4.00 , \text{atm} = 1.20 , \text{atm}$
Partial pressure of argon:
$P_{\text{Ar}} = 0.700 \times 4.00 , \text{atm} = 2.80 , \text{atm}$

Answer:

Partial pressure of helium ( $P_{\text{He}}$ ): 1.20 atm
Partial pressure of argon ( $P_{\text{Ar}}$ ): 2.80 atm

Question 8: What is the partial pressure of the remaining air in a blast furnace?

Given:

Total air pressure: $P_{\text{total}} = 0.99 , \text{atm}$
Partial pressure of carbon dioxide: $P_{\text{CO}_2} = 0.05 , \text{atm}$
Partial pressure of hydrogen sulfide: $P_{\text{H}_2\text{S}} = 0.02 , \text{atm}$

Solution:

Using Dalton’s Law of Partial Pressures:

$P_{\text{total}} = P_{\text{remaining air}} + P_{\text{CO}2} + P{\text{H}_2\text{S}}$

Rearranging to find $P_{\text{remaining air}}$ :

$P_{\text{remaining air}} = P_{\text{total}} - (P_{\text{CO}2} + P{\text{H}_2\text{S}})$

Substitute values:
$P_{\text{remaining air}} = 0.99 - (0.05 + 0.02)$
$P_{\text{remaining air}} = 0.99 - 0.07$

$P_{\text{remaining air}} = 0.92 , \text{atm}$

Answer: The partial pressure of the remaining air is 0.92 atm.

Question 9: What is the pressure, in mmHg, exerted by only the hydrogen gas?

Given:

Total pressure: $P_{\text{total}} = 753 , \text{mmHg}$
Water vapor pressure at $20.0^\circ\text{C}$ : $P_{\text{H}_2\text{O}} = 17.5 , \text{mmHg}$

Solution:

The pressure of hydrogen gas (

$P_{\text{H}2}$

) is calculated by subtracting the water vapor pressure from the total pressure:

$P{\text{H}2} = P{\text{total}} - P_{\text{H}_2\text{O}}$

Substitute values:

$P_{\text{H}2} = 753 - 17.5$

$P{\text{H}_2} = 735.5 , \text{mmHg}$

Answer: The pressure of hydrogen gas is 735.5 mmHg.

Question 10: How much space will the dry hydrogen gas sample occupy at STP?

Given:

Initial pressure of dry hydrogen gas: $P_1 = 735.5 , \text{mmHg}$
Initial volume: $V_1 = 138 , \text{mL}$
STP conditions:
- $P_2 = 760 , \text{mmHg}$
- $T_2 = 273.15 , \text{K}$
Initial temperature: $T_1 = 20.0^\circ\text{C} = 293.15 , \text{K}$

Solution:

Using the combined gas law:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Rearranging to solve for $V_2$ :

$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$

Substitute values:
$V_2 = \frac{735.5 \times 138 \times 273.15}{760 \times 293.15}$
$V_2 = \frac{27,678,701}{222,794}$

$V_2 = 124.2 , \text{mL}$

Answer: The dry hydrogen gas will occupy 124.2 mL at STP.

Question 11: Give the volume of dry gas at STP.

Given:

Initial pressure: $P_1 = 102.9 , \text{kPa}$
Water vapor pressure at $21^\circ\text{C}$ : $P_{\text{H}_2\text{O}} = 2.49 , \text{kPa}$
Dry gas pressure: $P_{\text{dry gas}} = P_{\text{atm}} - P_{\text{H}_2\text{O}} = 102.9 - 2.49 = 100.41 , \text{kPa}$
Initial volume: $V_1 = 325 , \text{mL}$
STP conditions:
- $P_2 = 101.3 , \text{kPa}$
- $T_2 = 273.15 , \text{K}$
Initial temperature: $T_1 = 21^\circ\text{C} = 294.15 , \text{K}$

Solution:

Using the combined gas law:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Rearranging to solve for $V_2$ :

$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$

Substitute values:
$V_2 = \frac{100.41 \times 325 \times 273.15}{101.3 \times 294.15}$
$V_2 = \frac{8,900,413.88}{29,768.895}$

$V_2 = 299.1 , \text{mL}$

Answer: The volume of dry gas at STP is 299.1 mL.

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