A 23 L flask at 254 K and 19 Torr contains nitrogen. What mass of nitrogen is present? Answer in units of g.

Use PV = nRT and solve for n. Then

n = grams/molar mass. Solve for grams.

To determine the mass of nitrogen present in the flask, we can use the ideal gas law equation:

PV = nRT

where:
P = pressure in atmospheres (since 1 Torr = 1/760 atm)
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin

First, let's convert the given pressure from torr to atm by dividing it by 760:
19 Torr / 760 Torr/atm = 0.025 atm

Next, let's convert the given temperature from Kelvin to Celsius:
254 K - 273.15 = -19.15 °C

Now, let's convert the volume from liters to moles by dividing it by the molar volume of a gas at STP (standard temperature and pressure), which is approximately 22.4 L/mol:
23 L / 22.4 L/mol ≈ 1.027 moles

Now, we can rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

n = (0.025 atm) * (1.027 moles) / (0.0821 L·atm/(mol·K)) * (-19.15 °C + 273.15 K)
n ≈ 0.001008 moles

Finally, to calculate the mass of nitrogen, we need to multiply the number of moles by its molar mass. Since the molar mass of nitrogen (N2) is approximately 28 g/mol:

mass = n * molar mass
mass = 0.001008 moles * 28 g/mol

Therefore, the mass of nitrogen in the flask is approximately 0.0282 g.

.2757g

I don't get that.

PV = nRT so
n = PV/RT
P = 19/760
V = 23
R = 0.08206
T = 254
and I find about 0.027 but you need to do it more accurately.
Then n = grams/molar mass
so grams = n from above x about 28 = about 0.8 gram.