434,620 views
18 votes
18 votes
You have 750 grams of Cr(S03). How many particles are present?

User Clarkk
by
2.7k points

1 Answer

16 votes
16 votes

Step-by-step explanation:

Before we find the number of particles we have to convert from grams to moles. To do that we can use the molar mass of our compound.

molar mass of Cr = 52.00 g/mol

molar mass of S = 32.07 g/mol

molar mass of O = 16.00 g/mol

molar mass of CrSO₃ = 1 * 52.00 g/mol + 1 * 32.07 g/mol + 3 * 16.00 g/mol

molar mass of to c = 132.07 g/mol

mass of onver = 750 g

moles of tles = 750 g/(132.07 g/mol)

moles of we ha = 5.68 molesvles we haCrSO₃CrSO₃CrSO₃CrSO₃

According to Avogadro's Number we have 6.022 * 10^23 particles in 1 mol of molCr. Let's use that relationship to find the answer to our problem.

1 mol of SO₃Cr = SO₃6.022 * 10^23 partic

n° of particles = 5.68 moles * 6.022 * 10^23 particles/(1 mol)

n° of particles = 3.42 * 10^24 particles

Answer: there are 3.42 * 10^24 particles.

User Lusid
by
2.6k points