When `0.004 M Na_(2)SO_(4)` is an isotonic acid with `0.01 M `glucose, the degree of dissociation of `Na_(2)SO_(4)` is

Options

A. 0.75
B. 0.5
C. 0.25
D. 0.85

Answer

A

Explanation

`0.004 M NaSO_4` solution is isotonic with 0.01 M solution of glucose, so their osmotic pressures are equal to each other. Osmotic pressure of 0.01 M glucose `(pi)_("glucose") =CST`
`therefore ` C = Concentration of solution = 0.01 M and S = Solution constant = 0.0821 L-atm/K/mol and T = A bsolute temperature
`therefore pi_("glucose") =0.01 xx0.0821 xxT ......(i) or pi _("glucose") =pi_(Na_2SO_4) "as" Na_2SO_4` is present in ionic state in solution
so,

`(pi_cal) Na_2SO_4 =C xxS xxT =0.004 xx0.0821 xxT ..........(ii)`
By van't Hoff factor `((pi_obs)Na_2SO_4)/((pi_cal)Na_2SO_4) =("number of particles after dissocation")/("number of particles before dissociation")=(1-alpha+2alpha+alpha)/1`
`(pi_obs)Na_2SO_4= pi_("glucose")`
`therefore (0.01xx0.0821xxT)/(0.004xx0.0821xxT)=(1+2alpha)/(1)or (10)/4 =(1+2alpha)/1or 10 =4 +8a and alpha=(10-4)/8 =0.75` and parcentage of = 75%