Gaseous hydrogen has a density of 0.0899 g/L at 0 °C, and gaseous chlorine has a density of 3.214 g/L at the same temperature. How many liters of each would you need if you wanted 1.0078 g of hydrogen and 35.45 g of chlorine?
Gaseous hydrogen has a density of 0.0899 g/L at 0 °C, and gaseous chlorine has a density of 3.214 g/L at the same temperature. How many liters of each would you need if you wanted 1.0078 g of hydrogen and 35.45 g of chlorine?
Chapter5: Errors In Chemical Analyses
Section: Chapter Questions
Problem 5.13QAP
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![**Chemistry Problem Example: Gas Density and Volume Calculations**
**Problem Statement:**
Gaseous hydrogen has a density of 0.0899 g/L at 0°C, and gaseous chlorine has a density of 3.214 g/L at the same temperature. How many liters of each would you need if you wanted 1.0078 g of hydrogen and 35.45 g of chlorine?
**Detailed Solution:**
To find the volume of gas needed, use the formula:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
For hydrogen:
1. Given:
- Density of hydrogen (\( \rho_H \)): 0.0899 g/L
- Mass of hydrogen (\( m_H \)): 1.0078 g
2. Applying the formula:
\[ V_H = \frac{m_H}{\rho_H} \]
\[ V_H = \frac{1.0078\text{ g}}{0.0899\text{ g/L}} \]
\[ V_H \approx 11.21\text{ L} \]
For chlorine:
1. Given:
- Density of chlorine (\( \rho_{Cl} \)): 3.214 g/L
- Mass of chlorine (\( m_{Cl} \)): 35.45 g
2. Applying the formula:
\[ V_{Cl} = \frac{m_{Cl}}{\rho_{Cl}} \]
\[ V_{Cl} = \frac{35.45\text{ g}}{3.214\text{ g/L}} \]
\[ V_{Cl} \approx 11.03\text{ L} \]
Hence, you would need approximately 11.21 liters of hydrogen and 11.03 liters of chlorine to meet the required amounts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e830786-058e-49dd-8798-047547f22476%2F932b0e09-3135-4314-be64-3d6f64fc3c69%2Fc5w1sa.png&w=3840&q=75)
Transcribed Image Text:**Chemistry Problem Example: Gas Density and Volume Calculations**
**Problem Statement:**
Gaseous hydrogen has a density of 0.0899 g/L at 0°C, and gaseous chlorine has a density of 3.214 g/L at the same temperature. How many liters of each would you need if you wanted 1.0078 g of hydrogen and 35.45 g of chlorine?
**Detailed Solution:**
To find the volume of gas needed, use the formula:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
For hydrogen:
1. Given:
- Density of hydrogen (\( \rho_H \)): 0.0899 g/L
- Mass of hydrogen (\( m_H \)): 1.0078 g
2. Applying the formula:
\[ V_H = \frac{m_H}{\rho_H} \]
\[ V_H = \frac{1.0078\text{ g}}{0.0899\text{ g/L}} \]
\[ V_H \approx 11.21\text{ L} \]
For chlorine:
1. Given:
- Density of chlorine (\( \rho_{Cl} \)): 3.214 g/L
- Mass of chlorine (\( m_{Cl} \)): 35.45 g
2. Applying the formula:
\[ V_{Cl} = \frac{m_{Cl}}{\rho_{Cl}} \]
\[ V_{Cl} = \frac{35.45\text{ g}}{3.214\text{ g/L}} \]
\[ V_{Cl} \approx 11.03\text{ L} \]
Hence, you would need approximately 11.21 liters of hydrogen and 11.03 liters of chlorine to meet the required amounts.
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