Question 6: Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of Cl₂ gas is 8.70 L at 895 torr and 24°C. (a) How many grams of Cl₂ are in the sample? ⚫ Atomic mass of CI = 35.453 g/mol • Molar mass of Cl₂ = 2 x 35.453 = 70.906 g/mol Solution: Use the Ideal Gas Law: Step 1: Convert Given Values • Pressure: P = 895 torr → atm PV= = nRT 1 P = 895 × = 1.1789 atm 760 • Temperature: Convert to Kelvin: T24273.15 = 297.15 K • Gas constant: R = 0.0821 L atm/mol. K Volume: V = 8.70 L Step 2: Solve for n . PV n = RT n = (1.1789)(8.70) (0.0821)(297.15) 10.25 n = = 0.420 mol 24.405 Step 3: Calculate Mass of Cl₂ Final Answer: 29.78 g of Cl₂. mass nx M mass= (0.420)(70.906) mass= 29.78 g (b) What volume will the Cl₂ occupy at standard temperature and pressure (STP)? STP Conditions: ⚫ P = 1.00 atm • T = 273.15 K Using Ideal Gas Law: Final Answer: 9.42 L at STP. nRT V = P (0.420) (0.0821) (273.15) V = 1.00 9.42 V = = 9.42 L 1.00 Final Summary of Answers: 1. Final volume of SO3: 25.0 mL 2. Final volume of the bubble: 0.0117 L 3. (a) Mass of Cl2: 29.78 g 4. (b) Volume of Cl₂ at STP: 9.42 L
Question 6: Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of Cl₂ gas is 8.70 L at 895 torr and 24°C. (a) How many grams of Cl₂ are in the sample? ⚫ Atomic mass of CI = 35.453 g/mol • Molar mass of Cl₂ = 2 x 35.453 = 70.906 g/mol Solution: Use the Ideal Gas Law: Step 1: Convert Given Values • Pressure: P = 895 torr → atm PV= = nRT 1 P = 895 × = 1.1789 atm 760 • Temperature: Convert to Kelvin: T24273.15 = 297.15 K • Gas constant: R = 0.0821 L atm/mol. K Volume: V = 8.70 L Step 2: Solve for n . PV n = RT n = (1.1789)(8.70) (0.0821)(297.15) 10.25 n = = 0.420 mol 24.405 Step 3: Calculate Mass of Cl₂ Final Answer: 29.78 g of Cl₂. mass nx M mass= (0.420)(70.906) mass= 29.78 g (b) What volume will the Cl₂ occupy at standard temperature and pressure (STP)? STP Conditions: ⚫ P = 1.00 atm • T = 273.15 K Using Ideal Gas Law: Final Answer: 9.42 L at STP. nRT V = P (0.420) (0.0821) (273.15) V = 1.00 9.42 V = = 9.42 L 1.00 Final Summary of Answers: 1. Final volume of SO3: 25.0 mL 2. Final volume of the bubble: 0.0117 L 3. (a) Mass of Cl2: 29.78 g 4. (b) Volume of Cl₂ at STP: 9.42 L
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter12: The Laws Of Thermodynamics
Section: Chapter Questions
Problem 72AP
Related questions
Question

Transcribed Image Text:Question 6:
Chlorine is widely used to purify municipal water supplies and to treat swimming pool
waters. Suppose that the volume of a particular sample of Cl₂ gas is 8.70 L at 895 torr
and 24°C.
(a) How many grams of Cl₂ are in the sample?
⚫ Atomic mass of CI = 35.453 g/mol
• Molar mass of Cl₂ = 2 x 35.453 = 70.906 g/mol
Solution:
Use the Ideal Gas Law:
Step 1: Convert Given Values
• Pressure: P = 895 torr → atm
PV=
= nRT
1
P = 895 ×
= 1.1789 atm
760
•
Temperature: Convert to Kelvin:
T24273.15 = 297.15 K
• Gas constant: R = 0.0821 L atm/mol. K
Volume: V = 8.70 L
Step 2: Solve for n
.
PV
n =
RT
n =
(1.1789)(8.70)
(0.0821)(297.15)
10.25
n =
= 0.420 mol
24.405
Step 3: Calculate Mass of Cl₂
Final Answer: 29.78 g of Cl₂.
mass nx M
mass=
(0.420)(70.906)
mass=
29.78 g

Transcribed Image Text:(b) What volume will the Cl₂ occupy at standard temperature and
pressure (STP)?
STP Conditions:
⚫ P = 1.00 atm
•
T = 273.15 K
Using Ideal Gas Law:
Final Answer: 9.42 L at STP.
nRT
V
=
P
(0.420) (0.0821) (273.15)
V =
1.00
9.42
V =
= 9.42 L
1.00
Final Summary of Answers:
1. Final volume of SO3: 25.0 mL
2. Final volume of the bubble: 0.0117 L
3. (a) Mass of Cl2: 29.78 g
4. (b) Volume of Cl₂ at STP: 9.42 L
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