Respuesta :
Answer:
As per the given statement:
[tex]V = 74.5mL[/tex]
[tex]P = 0.933 atm[/tex]
[tex]T = 30^{\circ} C[/tex]
Using ideal gas law equation:
[tex]PV = nRT[/tex]
where
P represents the Pressure of a gas
V represents the Volume it occupies
R represents the Universal constant , usually given as:
R = 0.082 atm L/ mol k
First convert Celsius into Kelvin.
T = 273 + 30 = 303 K
Use conversion:
1 L = 1000 mL
V = 74.5 mL = [tex]74.5 \times 10^{-3}[/tex] L
Now substitute the given values we have;
[tex]n = \frac{PV}{RT}[/tex]
[tex]n = \frac{0.933 \cdot 74.5 \cdot 10^{-3}}{0.082 \cdot 303}[/tex]
Simplify:
n = 0.00279757305
or
n ≈0.0028 moles.
Since carbon monoxide, CO has molar mass of 28.01 g/mol
then;
[tex]m = 0.0028 \times 28.01 = 0.078[/tex] g
Therefore, 0.078 g of CO(g) are there in 74.5 mL of the gas at 0.933 atm and 30 degree C
