Question

Sodium carbonate contains 43.4% Na 11.3% C and 45.3% O. If the law of constant composition is valid then weight of Na in 6 g of Na2CO3 obtained from other source Will be

Answer 1

Composition, Na = 43.4%, C = 11.3%, 0 = 43.3%

calculate it’s empirical ... Na = 43.4% C = 11.3. O = 43.3% relative number of moles of Na = 43.4/23 = 1.88.

It has empirical formula = Na2CO3 ...

A balloon contains 10 moles of a gas in a volume of 5 lts at 27 degree Celsius.

Answer 2

We will use the Unitary Method to solve the answer of this question.

Let the mass of the sodium carbonate originally is 100 g.

∴ 100 g of Sodium Carbonate contains 43.4 g of Na.

∴ 1 g of the ---------------------------------43.4/100 g of Na.

∴ 6 g of the ------------------------------ 43.4/100 × 6 g of Na.

=  2.604 g.

Now, 2.604 is just the 43.4 % of the 6 g of the compound. This proves that the Law of Constant Proportion is accurate.

This is the required answer.