Chapter 10, Problem 202A

Interpretation Introduction

Interpretation:

An experiment must be designed to determine the amount of water in alum.

Concept introduction:

Alums are mixed salt of aluminium and other metals which contains few molecules of water as water of crystallization.

Answer to Problem 202A

Some amount of alum (KAl (SO₄)_2.xH₂O) is weighed in a pre-weighed crucible and the total mass of the crucible and alum is noted.

Now the crucible is strongly heated to remove the water of crystallization completely.

Then dry alum salt and crucible is weighed.

Explanation of Solution

Let’s consider mass of crucible is x g and mass of hydrated salt y g.

Total mass before heating is (x+y) g.

After heating the mass of crucible and salt is z g.

Then mass of water

$\begin{array}{l}\text{=(Total mass before heating-total mass after heating)}\\ \text{=(x+y-z)}\text{.}\end{array}$

Mass of anhydrous alum $\text{=(z-x) g}\text{.}$

From the mole ratio of anhydrous salt and water the value of x in KAl (SO₄)₂.xH₂O can be found.

Conclusion

Amount of water is found by calculating weight loss in heating experiment.