JEE Main 2016 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

44 g of gypsum (CaSO4·2H2O, M=172) is first heated to 393 K and then to 773 K. The masses of the solid left after the first and the second heating…

Single correct option

3.44\ \mathrm{g}

of gypsum (

\mathrm{CaSO_4\cdot2H_2O}

M=172

) is first heated to

393\ \mathrm{K}

and then to

773\ \mathrm{K}

. The masses of the solid left after the first and the second heating are respectively (

\mathrm{CaSO_4}=136

)

Show answer and solution

Answer: (D) $2.90\ \mathrm{g}$ and $2.72\ \mathrm{g}$

Gypsum:

3.44/172=0.020

mol. At

393

K it loses

1\tfrac12

water molecules to give plaster of Paris

\mathrm{CaSO_4\cdot\tfrac12H_2O}

(

M=145

0.020\times145=2.90

g. On heating further to about

773

K the remaining water goes and anhydrous 'dead burnt'

\mathrm{CaSO_4}

is left:

0.020\times136=2.72

g. So the masses are

2.90

g and then

2.72

g, always decreasing; any option where the mass rises or stays at

3.44

g is eliminated.

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