Assuming 100% ionisation of $B a C l_{2} (a q)$ , what will be the value of van't Hoff factor?

1.5

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3

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2

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4

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Solution

The correct option is B $3$
$van't Hoff factor (i) = \frac{Number of molecules (actual)}{Number of molecules (expected)}$

For 100% ionisation, van't Hoff factor is equal to the number of particles in which 1 molecule of electrolyte dissociates.

$B a C l_{2} \to B a^{+ 2} + 2 C l^{-}$
$i = 3$

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