Question

in an aqueous solution kcl undergoes complete dissociation calculate osmotic pressure of 0.525Mkcl solution at 300k(R=0.0825latm/k)​

Answer 1

Answer: 25.86

Explanation:

Here vant Hoff factor i=2

Formula pi=CRT×i

=0.525×0.0821×300×2

=25.86

Answer 2

Solution:

π = iMRT

π => Osmotic Pressure

i => Van't Hoff Factor

M => Molarity

R => Universal Gas Constant

T => Temperature

M = 0.525

R = 0.0825 l.atm/k

T = 300 K

i = 2

i i.e. Van't Hoff factor of KCl will be 2 because we know that KCl dissociates into two ions i.e. K⁺ and Cl⁻

π = iMRT

π = 2 * 0.525 * 0.0825 * 300

π = 2 * 0.525 * 24.75

π = 2 * 12.99

π = 25.98

The osmotic pressure will be 25.98 atm(atmospheres)