Question

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1.) How many miles are there in 5g of Calcium ?

2.) Find the mass of 4 moles of Aluminium atom .

3.) Find the no. of atoms in 0.2 moles of Sodium .

4.) Find the moles in 9.033 × 10²⁴ atoms of helium .

5.) Find the no. of iron atoms in a piece of iron weighting 2.8g .


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Answer 1

Answer:

How many miles are there in 5g of Calcium ?

=Molar mass of Calcium=40.078 g/mol. So numbers of moles =5.00/40.078=0.125 mol.

Find the mass of 4 moles of Aluminium atom .

Mass of 4 moles of aluminium atoms = (27 x 4) = 108 g.

Find the no. of atoms in 0.2 moles of Sodium .

Hence, in 0.2 mole of Na(Sodium) there will be 1.2044×10^23 atoms

Find the moles in 9.033 × 10²⁴ atoms of helium .

9.033 × 10²⁴ atoms of helium are 15 moles of atoms.

Find the no. of iron atoms in a piece of iron weighting 2.8g .

We know that 1 mole of iron element contains 6.022×1023 atoms of iron. Thus, a piece of iron metal having a mass of 2.8 grams contains 3.011×1022 atoms of iron.

Explanation:

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Answer 2

Answer:

☢QuEsTiOn:

• 1.) How many miles are there in 5g of Calcium ?

• 2.) Find the mass of 4 moles of Aluminium atom .

• 3.) Find the no. of atoms in 0.2 moles of Sodium .

• 4.) Find the moles in 9.033 × 10²⁴ atoms of helium .

• 5.) Find the no. of iron atoms in a piece of iron weighting 2.8g .

☢AnSwEr:

• 1) Molar mass of Calcium=40.078 g/mol.

No. of moles =5.00/40.078=0.125 mol.

• 2) Mass of 1 mole of Aluminium atoms and the molar mass of aluminium is 27 g.

Mass of 4 moles of Aluminium atoms will be (4 x 27) g = 108 g

• 3) Symbol of Avogardo number of Atoms.

$\sf{ = 6.023 \times {10}^{23}atoms \times 0.2 }$

$\sf{ = 12.046 \times {10}^{23} atoms}$

Therefore, 0.2 moles of sodium (Na) contains 12.046 × 10²³ atoms.

• 4) find the sum of all no. of moles and all no. of atoms

$\sf{ = \frac{9.033 \times {10}^{24} }{6.022 \times {10}^{23} } }$

$\sf{ = \frac{9.033 \times {10}^{24 - 23} }{6.022} }$

$\sf{ = \frac{9.033 \times 10}{6.022} }$

$\sf{ = \frac{90.33}{6.022} }$

$\sf{ = \frac{90330}{6022} = 15. }$

Therefore, atoms of helium is 15.

• 5) Given that, weight of iron =2.8 g

Atomic mass = 56

Then,

$\sf{ = \frac{2.8}{56} }$

$\sf{1 mole = 6.022×10²³ atoms}$

$\sf{0.05 moles =6.022 ×0.05×10²³ Atoms}$

$\boxed{= 3.0115×10²² atoms.}$

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