to do the calculation, recall what density means. it is a measure of the amount of mass in a certain volume…

to do the calculation, recall what density means. it is a measure of the amount of mass in a certain volume. for gases, we usually measure density in g/l. so, density = grams / volume = 16.04 g / 32.333 l = 0.496 g/l if you try to do this problem again (from the beginning), using n = 2 instead of n = 1, you will find that you get the same answer. (notice that your final answer was reported with three significant figures.) now try to do each of the following problems. for each problem, use the space provided to develop your strategy, do your conversions, and calculate your answer. problem 4.54. calculate the density of carbon dioxide gas (co2) at a temperature of 50° c and a pressure of 850 torr. problem 4.55. calculate the density of oxygen (o2) at a temperature of 30° c and a pressure of 760 torr.

to do the calculation, recall what density means. it is a measure of the amount of mass in a certain volume. for gases, we usually measure density in g/l. so, density = grams / volume = 16.04 g / 32.333 l = 0.496 g/l if you try to do this problem again (from the beginning), using n = 2 instead of n = 1, you will find that you get the same answer. (notice that your final answer was reported with three significant figures.) now try to do each of the following problems. for each problem, use the space provided to develop your strategy, do your conversions, and calculate your answer. problem 4.54. calculate the density of carbon dioxide gas (co2) at a temperature of 50° c and a pressure of 850 torr. problem 4.55. calculate the density of oxygen (o2) at a temperature of 30° c and a pressure of 760 torr.

Answer

Explanation:

Step1: Recall ideal - gas law and density formula

The ideal - gas law is $PV = nRT$, and density $\rho=\frac{m}{V}$. Since $n=\frac{m}{M}$ (where $M$ is molar mass), we can rewrite the ideal - gas law in terms of density. Rearranging $PV = nRT$ gives $V=\frac{nRT}{P}$, and substituting $n=\frac{m}{M}$ into it, we get $V=\frac{mRT}{MP}$. Then $\rho=\frac{m}{V}=\frac{MP}{RT}$.

Step2: Solve for density of $CO_2$ in Problem 4.54

The molar mass of $CO_2$, $M_{CO_2}=(12.01 + 2\times16.00)\text{ g/mol}=44.01\text{ g/mol}$. The temperature $T = 50^{\circ}C=(50 + 273.15)\text{ K}=323.15\text{ K}$, and the pressure $P = 850\text{ torr}=\frac{850}{760}\text{ atm}\approx1.118\text{ atm}$. The ideal - gas constant $R = 0.0821\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$. Using $\rho=\frac{MP}{RT}$, we have $\rho_{CO_2}=\frac{44.01\text{ g/mol}\times1.118\text{ atm}}{0.0821\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times323.15\text{ K}}\approx1.86\text{ g/L}$.

Step3: Solve for density of $O_2$ in Problem 4.55

The molar mass of $O_2$, $M_{O_2}=2\times16.00\text{ g/mol}=32.00\text{ g/mol}$. The temperature $T = 30^{\circ}C=(30 + 273.15)\text{ K}=303.15\text{ K}$, and the pressure $P = 760\text{ torr}=1\text{ atm}$. Using $\rho=\frac{MP}{RT}$, we have $\rho_{O_2}=\frac{32.00\text{ g/mol}\times1\text{ atm}}{0.0821\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times303.15\text{ K}}\approx1.29\text{ g/L}$.

Answer:

Problem 4.54: The density of $CO_2$ is approximately $1.86\text{ g/L}$. Problem 4.55: The density of $O_2$ is approximately $1.29\text{ g/L}$.