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3 years ago

How many lead (Pb) atoms are in a 5.32 g sample of pure lead?

Chemistry

1 answer:

Elanso [62]3 years ago

4 0

Answer:

1.55×10²² molecules.

Explanation:

We'll begin by calculating the number of mole in 5.32 g of pure lead (Pb). This can be obtained as follow:

Mass of Pb = 5.32 g

Molar mass of Pb = 207 g/mol

Mole of Pb =?

Mole = mass /molar mass

Mole of Pb = 5.32/207

Mole of Pb = 0.0257 mole

Finally, we shall determine the number of molecules in 0.0257 mole of Pb. This can be obtained as follow:

From Avogadro's hypothesis,

I mole of Pb contains 6.02×10²³ molecules.

Therefore, 0.0257 mole will contain = 0.0257 × 6.02×10²³ = 1.55×10²² molecules.

Therefore, 5.32 g of pure lead (Pb) contains 1.55×10²² molecules.

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Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 5.90 moles of magnesium perchlorate, Mg(ClO4)2.

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5.90, 11.8, 47.2

Explanation:

Let’s remove the parentheses and write the formula as MgCl₂O₈.

We see that 1 mol Mg(ClO₄)₂ contains 1 mol Mg atoms, 2 mol Cl atoms, and 8 mol O atoms.

∴ $\text{Moles of Mg atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{1 mol Mg atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{5.90 mol Mg atoms}$

$\text{Moles of Cl atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{2 mol Cl atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{11.8 mol Cl atoms}$

$\text{Moles of O atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{8 mol O atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{47.2 mol O atoms}$

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3 years ago