Question

0.1 normal solution of k4 Fe CN 6 is ionized 80% find its osmotic pressure at 300 Kelvin​

Answer 1

Answer:

Explanation:

We have, molarity = n/V

= 0.1 mol/ litre

= 0.1/10^(-)3 mol /m³

= 102 mol/m³

As, πN × V = n ST

Or, πN = (n/V) ST

= 102 × 8.314 × 300 N / m²

Let degree of association be α

Therefore, for K4Fe(CN)6 ↔ 4K(+) + Fe(CN)^(4-)6

                                     1 - α   4α   α

It is given that α = 0.5

And,

π exp / πN = 1 + 4α

Therefore, π exp = πN (1 + 4α) =

= 10² × 8.314 × 300 × (1 + 4 × 0.5)

= 10² × 8.314 × 300 × (1 + 2)

Osmotic pressure = 7.483 × 105 N / m²

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