Question

the freezing point of a solution containing 5 gram of benzoic acid (M=122 g) in 35 gram of benzene is depressed by 2.94 Kelvin what is the percentage dissociation of benzoic acid if it forms a dimer in solution .
(KF for benzene = 4.9 Kelvin kg mole inverse).

Answer 1

Answer:

99.6

Explanation:

Answer 2

The Percentage dissociation of benzoic acid is 97.6 %

• Now, given  Depression in freezing point T(b) = 2.94 K, Weight of solute = 5 gm, Weight of solvent = 35gm, Molar mass of solvent = 122g, Kf for benzene = 4.9 Kkg/mol
• Let the experimental molar mass of solute be x.
• We know,  T(b) = Kf * m ,                  where m is molality of the solution.
• Also we know  m = Weight of solute * 1000/Molar mass of solute * Mass of solvent
• Now this implies;                                                                                      T(b) = Kf * weight of solute * 1000 / Molar mass of solute * Mass of solvent
• From data, 2.94 = 4.9 * 5 * 1000/ x * 35
• x = 4.9 * 5 * 1000 / 2.94 * 35
• x = 24500 / 102.9  =  238.095 g/mol.
• Now consider the equation,   (C6H5COOH)2 ⇆ 2 C6H5COOH
• Let d be degree of dissociation of the solute , then we would have (1-d) mol of benzoic acid left in undissociated form and correspondingly d/2 as dissociated moles of benzoic acid at equilibrium.
• Hence, total number of moles of particles at equilibrium is :                  1 - d + d/2 = 1 - d/2
• We know Vant Hoff factor i, is total number of moles of particles at equilibrium.
• Also, i = Normal molar mass/ Abnormal molar mass                                                           i = 122 / 238.095       (since normal molar mass of benzoic acid is 122g/mol)
•  i = 0.512
• So, i = 1 - d/2 = 0.512,    d/2 = 1 - 0.512 = 0.488 ,  d = 2 * 0.488 = 0.976
• Therefore percentage degree of dissociation is 0.976 8 100 = 97.6%