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Standard XII

Chemistry

Solubility

What is the m...

Question

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, $K_{s p} = 9.1 \times 10^{- 6}$ ).

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Solution

Let s be the solubility of $C a S O_{4}$ .
$C a S O_{4} ⇌ C a^{2 +} S O_{4}^{2 -}$
$[C a^{2 +}] = [S O_{4}^{2 +}] = s$
$K_{s p} [C a^{2 +}] [S O_{4}^{2 -}] = s \times s = s^{2} = 9.1 \times 10^{- 6}$
$s = 0.003016 M$
The molar mass of $C a S O_{4}$ is $40 + 32 + 64 = 136 g$ .
The solubility in g/L is $0.003017 \times 136 = 0.410 g / L$ .
Hence, 0.41 g dissolves in 1 L.
1 g will dissolve in $\frac{1}{0.41} = 2.43 L$ .

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Q. Hard water required for an experiment is not available in a school laboratory. However, the following salts are available in the laboratory. Select the salts which may be dissolved in water to make it hard for the experiment :

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(3) calcium chloride
(4) potassium sulphate
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Hard water required for an experiment is not available in a school laboratory. However, following salts are available in the laboratory. Select the salts which may be dissolved in water to make it hard for the experiment

(1) Calcium Sulphate

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(4) Potassium Sulphate

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Q. A mixture of

100 m m o l

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O H^{-}

in resulting solution, respectively, are:

[Molar mass of

C a (O H)_{2}, N a_{2} S O_{4}

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Q. A sample of hard water contains

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What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp=9.1×10−6).

Let s be the solubility of CaSO4.CaSO4⇌Ca2+SO2−4[Ca2+]=[SO2+4]=sKsp[Ca2+][SO2−4]=s×s=s2=9.1×10−6s=0.003016MThe molar mass of CaSO4 is 40+32+64=136g.The solubility in g/L is 0.003017×136=0.410g/L.Hence, 0.41 g dissolves in 1 L.1 g will dissolve in 10.41=2.43L.

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, $K_{s p} = 9.1 \times 10^{- 6}$ ).