Question

A solution containing \( 1.9 g \) per \( 100 mL \) of \( KCl \left(M=74.5 g mol ^{-1}\right) \) is isotonic with a solution containing \( 3 g \) per \( 100 mL \) of urea \( \left(M=60 g mol ^{-1}\right) \). Calculate the degree of dissociation of \( KCl \) solution. Assume that both the solutions have same temperature.

Answer 1

We have given,

Solution containing 1.9 g per 100 ml of KCl is isotonic with solution containing 3g per 100ml of area 

\(\therefore \) Number of moles of area = \(\frac3{60}\)

= 0.05 mol.

it means

KCl solution contains = 0.05 moles particles in solution

Number of moles of KCl = \(\frac{1.9}{74.5}\)

= 0.026 mol

Initially

at equilibrium 

0.026 (1 - α) mol 

0.026 α mol  

0.026 α mol  (α = degree of dissociation)

total number of moles of particles in solution 

= 0.026 - α 0.026 + 0.026 α + 0.026 α

= 0.026 + 0.026 α

but KCl solution contain = 0.05 moles particles

\(\therefore \) 0.026 + 0.026 α = 0.05

0.026 α = 0.024

α = \(\frac{0.024}{0.026}\)

α = 0.92

Hence, degree of dissociation of KCl is 0.92