Question

Find the mass of a sample of O2 gas which occupies .95 L at 1.2 atm and 355 K. a) 1.2 g O2 b) 0.012 g O2 c) 0.62 g O2 d) 6.2 g O2

Answer 1

Answer: (d) 6.2 g O2.

Explanation: P = 1.2 atm

V = 0.95 L

n = PV / RT = (1.2 atm * 0.95 L) / (0.08206 L atm / mol K * 355 K) = 0.036 mol

molar mass of O2 = 32.00 g/mol

mass = n * molar mass = 0.036 mol * 32.00 g/mol = 6.2 g

Answer 2

c) 0.62 g O2.

Step-by-step explanation:

To find the mass of the O2 gas sample, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure of the gas (in atm)

V = volume of the gas (in liters)

n = number of moles of the gas

R = ideal gas constant (0.0821 L·atm/mol·K)

T = temperature of the gas (in Kelvin)

We are given:

V = 0.95 L

P = 1.2 atm

T = 355 K

First, let's rearrange the ideal gas law equation to solve for the number of moles (n):

n = PV / RT

Now, we can substitute the given values:

n = (1.2 atm * 0.95 L) / (0.0821 L·atm/mol·K * 355 K)

n ≈ 0.0437 mol

To calculate the mass of the O2 gas sample, we need to use the molar mass of O2, which is 32 g/mol.

Mass = n * molar mass

Mass = 0.0437 mol * 32 g/mol

Mass ≈ 1.3984 g

Therefore, the mass of the O2 gas sample is approximately 1.4 g.