Answer: There are 0.00054 moles of silane gas (SiH4) are present in 8.68 mL measured at 18 0C and 1.50 atm.
Explanation:
Given: Volume = 8.68 mL (1 mL = 0.001 L) = [tex]8.68 \times 10^{-3} L[/tex],
Temperature = [tex]18^{o}C = (18 + 273) K = 291 K[/tex],
Pressure = 1.50 atm
The ideal gas formula is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\1.50 atm \times 8.68 \times 10^{-3} L = n \times 0.0821 L atm/mol K \times 291 K\\n = \frac{1.50 atm \times 8.68 \times 10^{-3} L}{0.0821 L atm/mol K \times 291 K}\\= \frac{0.01302}{23.8911}\\= 0.00054 mol[/tex]
Thus, we can conclude that there are 0.00054 moles of silane gas (SiH4) are present in 8.68 mL measured at 18 0C and 1.50 atm.
