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The dead burn...

Question

The dead burnt plaster is:

C a S O_{4} . 2 H_{2} O

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C a S O_{4} . \frac{1}{2} H_{2} O

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C a S O_{4}

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2 C a S O_{4} . H_{2} O

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Solution

The correct option is B $C a S O_{4}$
When plaster of Paris is heated above 393 $K$ its water of crystallization is lost and the leftover anhydrous calcium sulphate $C a S O_{4}$ is called the dead burnt plaster.

Hence, option $C$ is correct.

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Similar questions

$C a S O_{4} .2 H_{2} O Δ 120^{\circ} C - ----- \to A - - + 3 H_{2} O . A$ is

C a S O_{4} \cdot 2 H_{2} O

C a s O_{4} \cdot \frac{1}{2} H_{2} O

M a S O_{4} \cdot 7 H_{2} O

M g S O_{4} \cdot H_{2} O

C a S O_{4}

393 K

C a S O_{4} \cdot 2 H_{2} O

C a S O_{4} \cdot \frac{1}{2} H_{2} O

C a S O_{4} \cdot 2 H_{2} O (s) \to C a S O_{4} \cdot \frac{1}{2} H_{2} O (s) + \frac{3}{2} H_{2} O (g)

Δ_{f} H^{⊖}

C a S O_{4} \cdot 2 H_{2} O (s), C a S O_{4} \frac{1}{2} H_{2} O (s), H_{2} O (g)

- 2021.0 k J m o l^{- 1}

- 1575.0 k J m o l^{- 1}

- 241.8 k J m o l^{- 1}

194.0

130.0

188.0 J K^{- 1} m o l^{- 1} .

R = 8.314 J K^{- 1} m o l^{- 1} = 0.0831 L b a r m o l^{- 1} K^{- 1}

1 K g

C a S O_{4} \cdot 2 H_{2} O (s)

^{- 1}

C a S O_{4} \cdot \frac{1}{2} H_{2} O (s)

The common name of the compound with the molecular formula CaSO₄.2H₂O is:

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