Question

13.4 g of a sample of unstable hydrated salt: Na2SO4. xH20 was found to contain 6.3g of water.
Determine the number of water of crystallization.

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Answer 1

Answer:

The number of water of crystallization is 7.

Explanation:

Molar mass of $Na_2SO_4.xH_2O$=142 + 18 x g/mol

In (142 + 18 x) g/mol of $Na_2SO_4.xH_2O$ = 18x g/mol of water

Then in 13.4 g of $Na_2SO_4.xH_2O$ = 6.3 g

$\frac{18x}{142+18x}\times 13.4=6.3 g$

x = 7

The number of water of crystallization is 7.

Answer 2

The number of water of crystallization in $Na_2SO_4.xH_2O$ is 7.

Explanation:

Decomposition of hydrated sodium sulphate is given by:

$Na_2SO_4.xH_2O\rightarrow Na_2SO_4+xH_2O$

Molar mass of $Na_2SO_4$ = 142.04 g/mol

According to stoichiometry:

(142.04+18x) g of $Na_2SO_4.xH_2O$ decomposes to give 18x g of $H_2O$

Thus 13.4 g of $Na_2SO_4.xH_2O$ decomposes to give=$\frac {18x }{(142.04+18x)}\times 13.4$ of $H_2O$

But it is given 13.4 g of a sample of unstable hydrated salt $Na_2SO_4.xH_2O$ was found to contain 6.3 g of water.

Thus we can equate the two equations:

$\frac{18x}{(142.04+18x)}\times 13.4=6.3$

$x=7$  

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