Answer:
1) 33.64 Liter will be volume of ammonia.
2) [tex] 2.409\times 10^{23} [/tex] molecules of NaCl are in 23.40 grams NaCl.
Explanation:
1) Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = [tex]1 atm[/tex] (at STP)
V = Volume of gas = ?
n = number of moles of gas = 1.50 mol
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 273,15 K
Putting values in above equation, we get:
[tex]V=\frac{nRT}{P}=\frac{1.50 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}[/tex]
V = 33.64 L
2) [tex]N=n\times N_A[/tex]
Where:
N = Number of particles / atoms/ molecules
n = Number of moles
[tex]N_A=6.022\times 10^{23} mol^{-1}[/tex] = Avogadro's number
We have:
Molar mass of NaCl = 58.5 g/mol
n = [tex]\frac{23.40 g}{58.5 g/mol}=0.4 mol[/tex]
[tex]N=0.4 mol\times 6.022\times 10^{23} mol^{-1}=2.409\times 10^{23} molecules[/tex]
[tex] 2.409\times 10^{23} [/tex] molecules of NaCl are in 23.40 grams NaCl.
