Question

Calculate the no.of gram atoms in 5.3 g of NH4cl

Answer 1

Answer:

Calculate the no.of gram atoms in 5.3 g of NH₄Cl.

The gram atoms in in 5.3 g of NH₄Cl is 0.0991 moles/gram atoms

Explanation:

The formula for calculating the gram atomic mass is:

  = Mass of the compound given / Relative atomic mass of the compound

Mass is 5.3 g

Relative molecular mass of in NH₄Cl is: 14 + 1 ×4 + 35.5 = 53.5

Therefore the gram atoms in 5.3 g of NH₄Cl is:

5.3g/ 53.5gmol⁻¹ = 0.0991 moles/gram atoms

Therefore the gram atoms in 5.3 g of NH₄Cl is 0.0991 moles/gram atoms

Answer 2

.

Moles of hydrogen in 500 cm³ of hydrogen

22.4 liters = 22400 cm³

If 22400 cm³ = 1 mole

Then 500cm³ = 1 mole × 500cm³/22400cm³

                      ≈ 0.02232 moles

We can apply Avogadro's number/constant to calculate the number of molecules in the 0.02232 molecules of hydrogen gas.

If 1 mole = 6.022 × 10²³ molecules

Then 0.02232 moles = 6.022 × 10²³ × 0.02232 moles/1 mole

                                = 1.3441104 ×10²³ molecules

Therefore there are 1.3441104 ×10²³ molecules of hydrogen in 500cm³ of hydrogen gas. ( which has been given as n)

Find the number of molecules in 25 cm³ of oxygen at STP using the same formula.

22400 cm³ = 1 mole

25 cm³ = 1 mole × 25 cm³/22400 cm³

            ≈ 0.00112 moles

Find number of molecules in 0.00112 moles of oxygen

If 1 mole = 6.022 × 10²³ molecules

Then 0.00112 moles = 6.022 × 10²³ × 0.00112 moles/1 mole

                                = 6.721 × 10²⁰ molecules

Therefore there are 6.721 × 10²⁰ molecules of oxygen in 25cm³ of oxygen gas.

You can express this number in terms of n in respect to the number of molecules of hydrogen:

If  1.3441104 ×10²³ = n

Then 6.721 × 10²³ = n × 6.721 × 10²⁰ / 1.3441104 ×10²³

                            = 5n

Therefore in terms of n, there are 5n molecules of oxygen in 25cm³ of oxygen.