QuestionMay 7, 2025

10. A pure sample of a gas has a density of 1.31g/L at 1.00 atm and 25.0^circ C What is the molar mass of this gas? Propose a possible identity of this gas.

10. A pure sample of a gas has a density of 1.31g/L at 1.00 atm and 25.0^circ C What is the molar mass of this gas? Propose a possible identity of this gas.
10. A pure sample of a gas has a density of 1.31g/L at 1.00 atm and 25.0^circ C What is the molar mass of
this gas? Propose a possible identity of this gas.

Solution
4.7(226 votes)

Answer

Molar mass = 32.03 \, g/mol, possible identity = O_2. Explanation 1. Use the Ideal Gas Law to relate density and molar mass The ideal gas law is PV = nRT. Rearrange it as n = \frac{m}{M}, where m is mass and M is molar mass. Substituting n into the equation gives PM = \frac{m}{V}RT. Since \frac{m}{V} is density (d), we get **M = \frac{dRT}{P}**. 2. Substitute known values into the formula Given: d = 1.31 \, g/L, R = 0.0821 \, L·atm/(mol·K), T = 25.0^{\circ}C = 298.15 \, K, P = 1.00 \, atm. Substituting: M = \frac{(1.31)(0.0821)(298.15)}{1.00}. 3. Perform the calculation M = \frac{(1.31)(24.452)}{1.00} = 32.03 \, g/mol. 4. Propose a possible identity of the gas A molar mass of approximately 32 \, g/mol suggests the gas could be oxygen (O_2).

Explanation

1. Use the Ideal Gas Law to relate density and molar mass<br /> The ideal gas law is $PV = nRT$. Rearrange it as $n = \frac{m}{M}$, where $m$ is mass and $M$ is molar mass. Substituting $n$ into the equation gives $PM = \frac{m}{V}RT$. Since $\frac{m}{V}$ is density ($d$), we get **$M = \frac{dRT}{P}$**.<br /><br />2. Substitute known values into the formula<br /> Given: $d = 1.31 \, g/L$, $R = 0.0821 \, L·atm/(mol·K)$, $T = 25.0^{\circ}C = 298.15 \, K$, $P = 1.00 \, atm$. Substituting:<br />$M = \frac{(1.31)(0.0821)(298.15)}{1.00}$.<br /><br />3. Perform the calculation<br /> $M = \frac{(1.31)(24.452)}{1.00} = 32.03 \, g/mol$.<br /><br />4. Propose a possible identity of the gas<br /> A molar mass of approximately $32 \, g/mol$ suggests the gas could be oxygen ($O_2$).
Click to rate:

Similar Questions