what is the mass of O2 in grams?

Question

In a mixture of Cl2 and O2 gases, the partial pressure of Cl2 is 0.200 atm. The total pressure is 0.495 atm. The temperature is 32.0oC. If the total volume of the gas mixture is 349 mL, what is the mass of O2 in grams?

Answer 1

Oxygen pressure=total pressure-cl2 pressure

PV= nRT

n=RT/PV

mass=n/MW

MW of oxygen = 16x2

Answer 2

the partial rigidity of O2 is 0.365-0.a hundred and fifty = 0.185 atm The form of moles of O2 is derived from the formula pV =nRT the place p is the rigidity, V the quantity, n form of moles, T absolute temperature T = t +273, R consistent so n = pV/RT R = 8.31 ;T= 273+32 = 305°ok; p= 0.185 *10^5 Pa; V= 0.000538 m^3 we come across n = 0.0039 moles As each and each mole of O2 weights 32g we got here upon m = 0.0039*32=

Answer 3

Total Pressure of Gas Mixture = Sum of Partial pressure of individual component
so;
0.495 =0.200+Partial pressure of Oxygen
so;
Pressure of Oxygen = 0.295 atm
according to Ideal gas Law for oxygen ;
PV = nRT
where
P (pressure) = 0.295 atm
V (volume) = 349 mL = 0.349 Liters
n = Number of moles of oxygen = ???
R = Constant = 0.08205 atm.Liters/mole.K (units shoud be consistent )
T = 32 deg. C = 305 K
now again according to ideal gas law;
n=PV/RT
so
number of moles of oxygen = (0.295 * 0.349) / (0.08205 * 305 ) = 0.00411 gm moles
Mass of oxygen = moles * molecular weight
= 0.00411 * 32 = 0.132 grams is Answer

Answer 4

Pressure Cl2 = 0.200 atm
Pressure O2 = 0.495 - 0.200 = 0.295 atm

PV = nRt

n = PV / Rt = (0.295 atm)(0.349 l) / (.0821 l atm/mol K)( 305 K)

n= 4.11 Exp -4 moles O2

Mass O2 = (4.11 Exp -4 moles O2) X (32 grams / mole) = 0.131 grams

Answer: 0.131 grams O2