Question

122 g molº
241.98 gmol
or
1-
122
241.98
1-0.504 = 0.496
2
or
x = 2 * 0.496 = 0.992
Therefore, degree of association of benzoic acid in benzene is 99.2%.
emple 2.13 0.6 mL of acetic acid (CH,COOH), having density 1.06 g mL-', is
dissolved in 1 litre of water. The depression in freezing point
observed for this strength of acid was 0.0205°C. Calculate the van't
Hoff factor and the dissociation constant of acid.
Solution Number of moles of acetic acid
0.6 mL * 1.06 g mL
60 g mol
= 0.0106 mol = n
0.0106 mol
Molality = -
= 0.0106 mol 1-1
10​

Answer 1

Answer:

B

Explanation:

m=(benzoic acid)

=1000KbW1ΔTbW2

=244 (twice of actual molecular weight)