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
B
36.5
C
0.365
D
150

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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