Aluminum chloride (AICl3) is used in refining petroleum and manufacturing rubber
and lubricants. How many formula units are in a 35.6 g sample of aluminum
chloride?

Respuesta :

Answer:

mass change to mol

molar mass of ALCL3 = 133.5

mol = molar mass/mass

= 133.5/35.6

= 3.75 mol

number of formula units

= 3.75 mol × 6.02×10²³

= 2.258×10²⁴

2.258×10²⁴ formula units are there in 35.6g aluminum chloride.

How to find formula units?

The formula units of any compound are determined by multiplying the moles of the given compound by the Avogadro number.

mass change to mol

molar mass of ALCL3 = 133.5

mol = molar mass/mass

= 133.5/35.6

= 3.75 mol

number of formula units

= 3.75 mol × 6.02×10²³

= 2.258×10²⁴

Avogadro's number, the number of units in one mole of any substance, is equal to 6.02214076 × 1023.

Learn more about formula units here: brainly.com/question/25109150

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