Question

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?

Standard Temperature and Pressure (STP) = 273 K and 1.0 atm
Convert grams to moles by dividing by molar mass of O2
a
0.449 liters
6
0.432 liters
C
0.418 liters
d.
0.406 liters

Answer 1

Answer: The volume of 0.640 grams of
`O_(2)` gas at Standard Temperature and Pressure (STP) is 0.449 L.

Step-by-step explanation:

Given: Mass of
`O_(2)` gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of
`O_(2)` (molar mass = 32.0 g/mol) is as follows.

`No. of moles = (mass)/(molar mass)\\= (0.640 g)/(32.0 g/mol)\\= 0.02 mol`

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

`PV = nRT\\1.0 atm * V = 0.02 mol * 0.0821 L atm/mol K * 273 K\\V = 0.449 L`

Thus, we can conclude that the volume of 0.640 grams of
`O_(2)` gas at Standard Temperature and Pressure (STP) is 0.449 L.