Question

For the coagulation of 500 ml of a sulphide sol, 5 ml of 0.2 M CaCl2 solution is required. The flocculation value of CaCl2 is

Answer 1

What is the value of van't Hoff factor when the given electrolyte undergoes ... 0.1M, 0.01M and 0.001M, respectively. The value of ... Molar mass of CaCl2 ,M =40+2 X 3 5.5 = 111 g mol– 1.

Answer 2

Given:

For the coagulation of 500 ml of a sulphide sol, 5 ml of 0.2 M CaCl2 solution is required

To find:

The flocculation value of CaCl2  

Solution:

We know that,

1 Molar solution contains 1 Mole in 1 Litre

So, here we have

0.2 Molar of CaCl₂ contains 0.2 mol/L  of CaCl₂

Therefore, we can also say

1 L of CaCl₂ contains 0.2 moles of CaCl₂

⇒ 1000 ml of CaCl₂ contains 0.2 moles of CaCl₂

∴ 5 ml of CaCl₂ contains,

=   $\frac{0.2}{1000} \times 5$

=   $2 \times 10^-^4 \times 5$

=   $10^-^3 \:moles$

=   $\bold{1 \:mmoles}$

Now,

$\boxed{\underline{Flocculation\:Value}}$ : The concentration or the number of millimoles of an electrolyte solution that is required to cause the coagulation of one litre of a colloidal solution is called the flocculation value of the colloidal solution.

According to the question, we have

If 500 ml of sol requires 1 mmoles for coagulation

Then 1 L or 1000 ml sol will require = $\frac{1}{500}\times 1000$ = $\bold{2\:mmoles}$

Thus, the flocculation value of CaCl₂ is 2 mmoles.

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