Question

How many moles are equivalent to 3.58x1023 formula units of ZnCl2? a. 0.555 mol ZnCl2 c. 0.621 mol ZnCl2 b. 1 mol ZnCl2 d. 0.595 mol ZnCl2

Answer 1

There are approximately d. 0.595 mol of ZnCl₂ in 3.58×10²³ formula units.

Step-by-step explanation:

Consider the Avogadro's Constant
`N_A` or equivalently,
`L`:

`N_A \approx \rm 6.02* 10^(23)\;mol^(-1)`.

In other words, there are
`\rm 6.02* 10^(23)\;mol^(-1)` constituent particles (Wikipedia) in each mole of a substance. In this case,

• Zinc chloride
  `\rm ZnCl_2` is the substance, and
• 
  `\rm ZnCl_2` formula units are the constituents.

`N = 3.58* 10^(23)`.

`\displaystyle n = (N)/(N_A) =\rm (3.58* 10^(23))/(6.02* 10^(23)\;mol^(-1)) = 0.595\;mol`.