Chemistry: How do you calculate the molarity of O2(oxygen) given that the pressure in bars of O2 is "0.20 bar"

Question

I have tried using the n=pv/rt but what else am i suppose to do inorder to arrive at the answer which is 0.25nM. Can someone please play with this a little bit for me so i could know what steps to take. thank you

Answer 1

Convert bar to atm 0.9869 [atm/bar] * .20 [bar] = .1974 atm

Molarity is in units of [moles/L], Using PV=nRT,

solve for n/V = P/RT which is in [moles/L]

n/V = (.1974[atm]) / (0.0821 [L*atm/mol*K])*(T [K])

Assuming T = 273.15 K, (standard temperature),

n/V = 0.0088 [moles/L] or 0.0088M

Answer 2

Are you assuming that your temperature is 273 K? IF that's the case, I'm getting a different answer.

What I did was solved the ideal gas law for n/V which should be concentration in mol/L or M. Starting with a pressure of 0.2 bar, divide that by 760 bar/atm to convert the pressure to atmospheres. Then divide the pressure by R*T. Using 0.0821 for R and 273K for temperature gives me 1.17 X 10*-5 M for the concentration, or 11.7 micromolar.

Unless the temperature given in your problem is really different, I'm not sure how you can get that concentration.

Answer 3

Bar To Atm Calculator

Answer 4

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