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5. How many atoms and molecules of sulphur are present in 64.0 g of sulphur (S 8 )?

1 Answer

5 votes

Answer:

There are
1.202* 10^(24) atoms and
1.502* 10^(23) molecules in the compound.

Step-by-step explanation:

The molar mass of the sulphur is
32.065\,(g)/(mol). The Avogradro's Law states that exists
6.022* 10^(23)\,(atom)/(mol). The quantity of atoms in a quantity of mass is derived from dividing the mass by the molar mass and multiplying it by the Avogadro's Number. That is:


n_(atom) = m_(S)\cdot (n_(A))/(M_(S))

Where:


m_(S) - Mass of the sample, measured in grams.


n_(A) - Avogadro's Number, measured in atoms per mole.


M_(S) - Molar mass of the sulphur, measured in grams per mole.

If
m_(S) = 64\,g,
n_(A) = 6.022* 10^(23)\,(atoms)/(mol) and
M_(S) = 32.065\,(g)/(mol), then:


n_(atom) = (64\,g)\cdot \left((6.022* 10^(23)\,(atoms)/(mol) )/(32.065\,(g)/(mol) )\right)


n_(atom) = 1.202* 10^(24)\,atoms

There are
1.202* 10^(24) atoms in the compound.

Now, the molecular weight of the compound is:


M_{S_(8)} = 8\cdot \left(32.065\,(g)/(mol) \right)


M_{S_(8)} = 256.52\,(g)/(mol)

The quantity of molecules in a quantity of mass is derived from dividing the mass by the molecular weight and multiplying it by the Avogadro's Number. That is:


n_(molecule) = m_{S_(8)}\cdot \frac{n_(A)}{M_{S_(8)}}

Where:


m_{S_(8)} - Mass of the sample, measured in grams.


n_(A) - Avogadro's Number, measured in atoms per mole.


M_{S_(8)} - Molecular weight of the compound (octosulphur), measured in grams per mole.

If
m_{S_(8)} = 64\,g,
n_(A) = 6.022* 10^(23)\,(molecules)/(mol) and
M_{S_(8)} = 256.52\,(g)/(mol), then:


n_(molecule) = (64\,g)\cdot \left((6.022* 10^(23)\,(molecules)/(mol) )/(256.52\,(g)/(mol) )\right)


n_(molecule) = 1.502* 10^(23)\,molecules

There are
1.502* 10^(23) molecules in the compound.

User Najim El Guennouni
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