What volume in liter will be occupied by 162.3 g of carbon monoxide at 273k and 760 torr?

Pv=nRT

n= 162.3/12

V=nRT/P

check the units on R to make certain you are consistent with K, Torr.

Please explain in detail where is 3 and 12 come from since you said n=163•3/12

Thank you

To find the volume of a gas, we can use the ideal gas law equation: PV = nRT, where P represents the pressure of the gas, V represents the volume, n represents the number of moles of gas, R represents the ideal gas constant, and T represents the temperature in Kelvin.

In this case, we are given the following information:
- Mass of carbon monoxide (CO): 162.3 g
- Temperature (T): 273 K
- Pressure (P): 760 torr

To calculate the volume (V), we need to find the number of moles (n) first. To do this, we can use the molar mass of carbon monoxide.

The molar mass of CO is calculated as follows:
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol

Therefore, the molar mass of CO = (12.01 g/mol) + (16.00 g/mol) = 28.01 g/mol

Now, we can find the number of moles (n) using the formula:
n = mass / molar mass

n = 162.3 g / 28.01 g/mol
≈ 5.79 mol

Now that we have the number of moles, we can proceed to calculate the volume (V). Rearranging the ideal gas law equation gives us the formula for volume:

V = (nRT) / P

Let's plug in the values:
V = (5.79 mol * 0.0821 L·atm/(mol·K) * 273 K) / 760 torr

Now, we need to convert torr to atm:
1 atm = 760 torr
So, 760 torr = 1 atm

V = (5.79 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
≈ 12.29 L

Therefore, the volume occupied by 162.3 g of carbon monoxide at 273 K and 760 torr is approximately 12.29 liters.