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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 :

(1) calcium sulphate
(2) sodium sulphate
(3) calcium chloride
(4) potassium sulphate
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(6) magnesium chloride

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

(2) Sodium Sulphate

(3) Calcium Chloride

(4) Potassium Sulphate

(5) Sodium Hydrogen Carbonate

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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}

and

C a S O_{4}

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and

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5.5 \times 10^{- 6}]

Q. A sample of hard water contains

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Q. A student requires hard water for an experiment in his laboratory which is not available in the neighbouring area. In the laboratory there are some salts, which when dissolved in distilled water can convert it into hard water. Select from the following groups of salts, which when dissolved in distilled water will make it hard.
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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, $K_{s p} = 9.1 \times 10^{- 6}$ ).