Question

S gram Na2co3 Is given calculate.

1.No. of moles of Na2co3.
2. No. of moles of Na and c. .
3. No. of Na atoms c and o atoms.


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Answer 1

Answer:

• Number of moles of Na₂CO₃ = 0.5 moles

• Number of moles of Na = 1 mole; Number of moles of C = 0.5 moles; Number of moles of O = 1.5 moles

• Number of atoms of Na = 6.022 × 10²³ atoms; Number of atoms of C = 3.011 × 10²³ atoms; Number of atoms of O =  9.033 × 10²³ atoms  

Explanation:

• We are given with 53 g of Na₂CO₃

• Let, us first calculate the molecular mass of 1 mole of Na₂CO₃

• → Gram molecular mass of Na₂CO₃ = 2 (gram atomic mass of Na) + 1 (gram Atomic mass of C) + 3 (gram atomic mass of O) = 2 (23) + 1(12) + 3(16) = 46 + 12 + 48 = 106 g

So,

• → number of moles Na₂CO₃ in 53 g = 53 / 106 = 0.5 moles

• Now, Since there are 2 moles of Na in 1 mole of Na₂CO₃ therefore,

• → number of moles of Na in 0.5 moles of Na₂CO₃ = 0.5 × 2 = 1 mole

• Similarly, Since there is 1 mole of C in 1 mole of Na₂CO₃ so,

• → number of moles of C in 0.5 moles of Na₂CO₃ = 0.5 × 1 = 0.5 moles

• And, there are 3 moles of O in 1 mole of Na₂CO₃ therefore,

• → number of moles of O in 0.5 moles of Na₂CO₃ = 0.5 × 3 = 1.5 moles

Further,

• Since one mole of a substance contains 6.022 × 10²³ units of that substance. ( n = 6.022 × 10²³ is called avogadro's constant )

so,

• → Atoms of Na in 0.5 moles of Na₂CO₃ = (number of moles of Na) × n = 1 × 6.022 × 10²³ = 6.022 × 10²³ atoms

• → Atoms of C in 0.5 moles of Na₂CO₃ = (number of moles of C) × n = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ atoms

• → Atoms of O in 0.5 moles of Na₂CO₃ = (number of moles of O) × n = 1.5 × 6.022 × 10²³ = 9.033 × 10²³ atoms.