A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO, H. Calculate the fraction of 4-chlorobutanoic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits. 1% x 5 ?

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

The ICE table is then shown as:

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

where ;

Since the value for is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

Dissociated form of  4-chlorobutanoic acid =

Percentage dissociated =

Percentage dissociated =

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole