Answer:
61.3 g/mol
Step-by-step explanation:
We can use the Ideal Gas Law to solve this problem:
pV = nRT
Since n = m/M, the equation becomes
pV = (m/M)RT Multiply each side by M
pVM = RT Divide each side by RT
M = (mRT)/(pV)
Data:
m = 0.675 g
R = 0.0.083 14 bar·L·K⁻¹mol⁻¹
T = 0 °C = 273.15 K
p = 1 bar
V = 250 mL = 0.250 L
Calculation:
M= (0.675 × 0.083 14 × 273.15)/(1 × 0.250)
M= 15.33/0.250
M= 61.3 g/mol
The molecular weight of the 250ml of gas found at STP has a molar mass of 60.81 g/mol.
According to the ideal gas equation:
PV= nRT
P = pressure = 1 atm at STP
V= volume = 250 ml = 0.25 L
n = moles
R = constant = 0.0821 atmp
T = temperature = 273.15 K
moles can be written as [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
PV = [tex]\rm \dfrac{weight}{molecular\;weight}\;\times\;RT[/tex]
1 [tex]\times[/tex] 0.25 = [tex]\rm \dfrac{0.675}{molecular\;weight}[/tex] [tex]\times[/tex] 0.0821 [tex]\times[/tex] 273.15
0.25 = [tex]\rm \dfrac{0.675}{molecular\;weight}[/tex] [tex]\times[/tex] 22.425
molecular weight = [tex]\rm \dfrac{22.425}{0.25}\;\times\;0.625[/tex]
molecular weight = 60.81 g/mol.
The molecular weight of the gas found at STP has a molar mass of 60.81 g/mol.
For more information about the molar mass, refer to the link:
https://brainly.com/question/12127540
