Question

Chemically, Plaster of Paris is calcium sulphate hemihydrate, i.e. containing half molecule of water of crystallisation. It is represented by the formula, ${\text{CaSO}}_4.\frac{1}{2}{\text{H}}_2\text{O}$. Half molecule of water of crystallisation means that one water molecule is shared by two formula units of $CaS{O}_4$. Hence we also represent its formula as $(CaS{O}_4{)}_2.H_{2}O$. The name, Plaster of Paris, was given to this compound because for the first time, it was made from gypsum which was mainly found in Paris. Gypsum has the chemical formula, $CaS{O}_4.2H_{2}O$. When plaster of Paris is mixed with water it sets into a hard mass.

(a) The difference in number of water molecules present as water of crystallisation in gypsum and Plaster of paris is
(i) 5/2 (ii) 2 (iii)1/2 (iv) 3/2

(b) Plaster of Paris hardens by
(i) giving off $C{O}_2$ (ii) changing into $CaC{O}_3$
(iii) combining with water (iv) giving out water

(c) Plaster of Paris is obtained by
(i) adding water to calcium sulphate
(ii) adding sulphuric acid to calcium hydroxide
(iii) heating gypsum to a very high temperature
(iv) heating gypsum to 373 K.
[3 Marks]

Answer 1

(a) (iv)3/2
Molecular formula of gypsum and Plaster of Paris respectively are $CaS{O}_4.2H_{2}O$ and $CaS{O}_4.\frac{1}{2}{H}_{2}O$
Hence the difference in number of water molecules present as water of crystallisation is
$2-\frac{1}{2}=\frac{3}{2}$
[1 Mark]
(b) (iii) combining with water
Plaster of Paris $({\text{CaSO}}_4.\frac{1}{2}{\text{H}}_2\text{O})$ is a white powder which on mixing with water sets into hard solid mass due to formation of gypsum$({\text{CaSO}}_4.{2H}_2\text{O})$.

${\text{CaSO}}_4.\frac{1}{2}{\text{H}}_2\text{O}+1\frac{1}{2}{\text{H}}_2\text{O(l)}\to {\text{CaSO}}_4.2{\text{H}}_2\text{O}$

[1 Mark]
(c) (iv) heating gypsum to 373 K.
Gypsum when heated at 373 K loses one and half molecules of water. The product formed is Plaster of Paris having chemical name calcium sulphate hemihydrate.

$CaS{O}_4.2H_{2}O\stackrel{373.K}{⟶}CaS{O}_4.\frac{1}{2}{H}_{2}O$ + $1\frac{1}{2}{H}_{2}O$
[1 Mark]