What mass of Oxygen gas would occupy a volume of 1800mL at Standard conditions? Atomic weight of oxygen is 16.

Md Ariful Islam

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Standart conditions: Temperature 273.15 K Pressure 101.325 kPa Using the universal gas law: pV = nRT or pV = mRT/M hence, m(O2) = pVM/RT where p - pressure V - volume M - molar mass of O2 R - the universal gas law constant of 8.3145 m^3*Pa/(mol*K) M(O2) = 2*16 = 32 g/mol m(O2) = 101325 Pa * 0.0018 m^3 * 32 g/mol / ( 8.3145 m^3*Pa/(mol*K) * 273.15 K) = 2.57 g
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