A sample of ammonia gas occupies 2.65 L when measured at 28 degrees celsius and 1.38 atm. What is the density of the ammonia under these conditions?
Please help, I'm not sure which formula to use! Thank you.
0.951g/L
I would use this one.
P*molar mass = density*RT
density will be in g/L.
Thank you! Did you get 1.26 g/L
No. The answer is less than 1 g/L.
I obtained 0.9498 which I would round to 0.950 g/L. I used 17 for the molar mass and 0.08206 for R.
To find the density of ammonia under these conditions, you can use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of the gas,
R is the ideal gas constant,
T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin. The Kelvin temperature scale is simply Celsius + 273.15. In this case, the temperature is 28 degrees Celsius, so the corresponding Kelvin temperature would be 28 + 273.15 = 301.15 K.
Now we have the pressure, volume, and temperature in the correct units. The ideal gas constant, R, is typically given as 0.0821 L·atm/(mol·K).
PV = nRT
Since we want to find the density, we'll rearrange the equation and solve for n/V:
n/V = (P/RT)
The density, ρ, is given by the equation:
ρ = (molar mass of the gas) x (n/V)
The molar mass of ammonia (NH3) is approximately 17.03 g/mol.
So, the final equation to find the density of ammonia is:
ρ = (17.03 g/mol) x (P/RT)
Using the given values: P = 1.38 atm, V = 2.65 L, R = 0.0821 L·atm/(mol·K), T = 301.15 K, we can substitute them into the equation to find the density.