Question

What is the value of x when 32.2g of na2so4xh2o are heated leaving 14.2g of anhydrous na2so4? Mr (h2o) = 18; mr (na2so4) = 142.

Answer 1

The value of x is 10

Explanation :

Since after heating it gives anhydrous Na₂SO₄ which means the difference in mass was the mass of water.

Mass of Na₂SO₄.xH₂O = 32.2

Mass of Na₂SO₄ = 14.2

=> mass of H₂O = 32.2 - 14.2 = 18

Molecular mass of water = 18

=> number of moles of water = 18/18 = 1

Mass of Na₂SO₄ = 14.2

Molecular mass of Na₂SO₄ = 142

=> number of moles of Na₂SO₄ = 14.2/142 = 0.1

Hence the ratio of number of moles = 0.1 : 1 = 1 : 10

In Na₂SO₄.xH₂O ratio of number of moles of = 1 : x

comparing both these ratio we get,

1/x = 1/10

=> x = 10

Hence the value of x is 10

Answer 2

Answer: The value of x is 10.

Explanation:

Molecular mass of $Na_2SO_4.xH_2O$ = 142+18x g

Molecular mass of $Na_2SO_4$ = 142 g

$Na_2SO_4.xH_2O\rightarrow Na_2SO_4+xH_2O$

Thus 142+18 x of  $Na_2SO_4.xH_2O$ on heating gives = 142 g of $Na_2SO_4$

32.2 g of $Na_2SO_4.xH_2O$ on heating should give = $\frac{142}{142+18x}\times 32.2$ g of $Na_2SO_4$ which is given to be equal to 14.2 g

Thus $\frac{142}{142+18x}\times 32.2= 14.2$

Solving for x

$x=10$