If you have 4.9 x 10^24 molecules of o2 at stp, how many liters of o2 do you have?

Given:

Molecules of O2 at STP = 4.9 * 10²⁴

STP conditions implies-

Temperature T = 273 K

Pressure = 1 atm

To determine:

The volume (V) of O2

Explanation:

Based on the ideal gas equation:

PV = nRT(1)

where n = number of moles of O2

R = gas constant = 0.0821 L.atm/mol-K

Now, 1 mole of O2 corresponds to 6.023*10²³ molecules

# moles of O2 corresponding to the given molecules=

= 4.9*10²⁴ molecules O2 * 1 mole O2/6.023*10²³ molecules = 8.135 moles

From equation(1) we have:

V=nRT/P = 8.135*0.0821*273/1 = 182.3 L

Ans: Volume of O2 under the given conditions is 182.3 L

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