The density of aluminum is 270 gml. How many aluminum atoms will be contained in a 1.57 Liter sample of the element?

9.45 × 10^21atoms

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

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the answer is c. nacl