Question

7.6 g KBr in 1250 ml solution was found to have an osmotic pressure of 1.804 atm at 27°C.
Calculate degree of ionization and Van't Hoff factor.​

Answer 1

Explanation:

degree of ionisation=44%

van't Hoff factor =1.44

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Answer 2

Given :

mass of KBr = 7.6 g

Volume = 1250 ml

Osmotic pressure = 1.804 atm

Temperature (T) = 27°C = 300 K

To find :

Degree of ionization and Van't Hoff factor.​

Solution :

In case of electrolytic solution, the formula for osmotic pressure is -

Osmotic Pressure (π) = iCRT

where, i = Vant Hoff's factor; C = Concentration; R = Molar gas constant; T = Temperature.

  Osmotic Pressure (π) = i × $\frac{7.6/(80 + 39)}{1.25}$ × 0.08 × 300

⇒      1.804                       = i × 1.226

⇒          i                           = $\frac{1.804}{1.226}$

∴           i                           = 1.5

∴ The Vant Hoff's factor is 1.5

We know, for dissociation,

Degree of dissociation ( $\alpha$ ) =  $\frac{i - 1}{n- 1}$        (n = number of particles dissociated)

                                             = $\frac{1.5 - 1}{2 - 1}$

                                             = 0.5

∴ Degree of dissociation is 0.5.