Respuesta :
Answer:
0.32 g O2
Explanation:
740 mmHg= 0.97 atm
25°C=298 K
250 mL=0.25 L
PV=nRT
n= PV/RT
n= (0.97 atm*0.25L)/(0.0821 * 298K)
n=9.91 x 10^-3 mol O2
1 mol O2 -> 32 g
9.91 x 10^-3 mol O2 -> x x= 0.32 g O2
The mass of the sample of oxygen gas with a pressure of 740 mm Hg has been 0.318 grams.
The sample has been assumed to be an ideal gas. According to the ideal gas equation:
PV = nRT
P = pressure = 740 mm Hg = 0.973 atm
V = volume = 250 ml = 0.25 L
n = moles of gas
R = constant = 0.0821
T = temperature = 25[tex]\rm ^\circ C[/tex] = 298 K
Substituting the values:
0.973 [tex]\times[/tex] 0.25 = n [tex]\times[/tex] 0.0821 [tex]\times[/tex] 298
0.24 = 24.465 n
n =0.0099 mol
The moles of the oxygen sample have been 0.0099 mol.
The molecular weight of oxygen gas = 32 g/mol
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
0.0099 = [tex]\rm \dfrac{weight}{32}[/tex]
Weight of oxygen gas = 0.0099 [tex]\times[/tex] 32 grams
Weight of oxygen gas = 0.318 grams.
The mass of the sample of oxygen gas with a pressure of 740 mm Hg has been 0.318 grams.
For more information about the mass of the sample, refer to the link:
https://brainly.com/question/2460334
