Answer:
\(6.018\times 10^{23} molecules\) of gas are present in 22.41 liter of gas at STP.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = 1 atm (STP)
V = Volume of gas = 22.41 L
n = number of moles of gas = ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 273.15 K (STP)
Putting values in above equation, we get:
\((1 atm)\times 22.41 L=n\times (0.0821L.atm/mol.K)\times 273.15 K\\\
=0.9993 mole\)
1 mole = \(N_A=6.022\times 10^{23} molecules/atoms\)
Number of molecules in 0.9993 moles of gas:
\(=0.9993 mol\times 6.022\times 10^{23} molecules=6.018\times 10^{23} molecules\)
\(6.018\times 10^{23} molecules\) of gas are present in 22.41 liter of gas at STP.
