9.45 × 10^21atoms
Step-by-step explanation:
Density = mass/volume
According to the question, Density = 270g/L, v = 1.57L
Hence,
270 = m/15.7
m = 4239g
mole = mass/molar mass
Molar mass of Aluminum = 27g/mol
mole = 4239/27
mole = 157mol
Using Avogadro's number, number of atoms of Aluminum =
157 × 6.02 × 10^23
= 945.14 × 10^23
= 9.45 × 10^21atoms