1. 1 mol=6.02.10²³ particles
[tex]\tt \dfrac{3.63\times 10^{23}}{6.02\times 10^{23}}=0.602~moles[/tex]
2. Molar mass = Total atomic mass of component
AlBr₃ = Ar Al + 3. Ar Br
AlBr₃ = 27 + 3. 80 = 267 g/mol
3. mass = mol x MW
MW MnCl₂ = 55 + 2.35.5=126 g/mol
mass = 31.6 x 126 =3981.6 g
4. at STP, 1 mol =22.4 L, so for 5.6 L :
[tex]\tt \dfrac{5.6}{22.4}=0.25~moles[/tex]
5. moles = mass : MW
MW CH₃OH = 12.1 + 4. 1 + 16.1 =32 g/mol
[tex]\tt moles=\dfrac{0.5}{32}=0.016[/tex]
6. 1 mol = 22.4 L at STP
1 mol = 6.02.10²³ particles
moles for 8.20 x 10¹⁴ molecules
[tex]\tt \dfrac{8.2\times 10^{14}}{6.02\times 10^{23}}=1.36\times 10^{-9}[/tex]
volume :
[tex]\tt 1.36\times 10^{-19}\times 22.4=3.05\times 10^{-8}~L[/tex]
