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Question

The coagulation of $$200$$ mL of a positive colloid took place when $$0.73$$ g $$HCl$$ was added to it without changing the volume much. The flocculation value of $$HCl$$ for the colloid is: 


A
100
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B
36.5
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C
0.365
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D
150
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Solution

The correct option is A $$100$$
200 mL of the sol required = 0.73 g $$HCl$$. 
The molecular weight of $$HCl$$ is $$36.5$$ g/mol.
$$=\dfrac{0.73 \; g}{36.5 \; g/mol} = 0.02 mol =20 mmol$$
$$\therefore$$ 1000 mL (1 L ) of the sol will require
$$=\dfrac{1000}{200}\times 20 =100 mmol$$
Thus, the flocculation value of the $$HCl$$ for the colloid is $$100$$.

Chemistry

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