Answer: The density of the given sample of hydrogen gas is 0.061 g/L
Step-by-step explanation:
Assuming ideal gas behavior, the equation follows:
PV = nRT
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://img.qammunity.org/2021/formulas/chemistry/college/e4lb9duyomysx0p41hk9jd8smtfdkqfqms.png)
Rearranging the above equation:
![P=(m)/(M)(RT)/(V)](https://img.qammunity.org/2021/formulas/chemistry/college/nqn0e9ayc6eqk83kz08si4l8brosj0fqmy.png)
We know that:
![\text{Density}=\frac{\text{Mass}}{\text{Volume}}](https://img.qammunity.org/2021/formulas/chemistry/college/hzer7715bnzribk2jcgmil7kwybeikd0nd.png)
Rearranging the above equation:
......(1)
We are given:
P = pressure of the gas = 0.799 atm
d = density of hydrogen gas = ?
R = Gas constant =
![0.0821\text{ L . atm }mol^(-1)K^(-1)](https://img.qammunity.org/2021/formulas/chemistry/college/mr9x4zq04qfdp4vsva0porweqtez3lpgz2.png)
T = temperature of the gas =
![47^oC=[47+273]K=320K](https://img.qammunity.org/2021/formulas/chemistry/college/7cnmypjdh02h1irgbw90ur8mcilqk46l8r.png)
M = molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
![0.799atm=\frac{d* 0.0821\text{ L.atm }mol^(-1)K^(-1)* 320K}{2g/mol}\\\\d=(0.799* 2)/(0.0821* 320)=0.061g/L](https://img.qammunity.org/2021/formulas/chemistry/college/tbx6u4ibpgltvoueoo6kr0gc484u7cuel2.png)
Hence, the density of the given sample of hydrogen gas is 0.061 g/L