Question
Fri September 02, 2011

# 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1C.Calculate the van't Hoff tion factor and dissocition constant of fluoroacetic acid.

Wed September 07, 2011

Number of moles of CH2FCOOH = 19.5 /78 =0.25

Molality (m) = (0.25 x 1000)/500 = 0.50 mol/Kg

Now,                Tf =Kf x m   =1.86 x 0.50

=0.93 K

Van't Hoff factor = (observed freezing point depression /calculated frezing point depression)

= 1/0.93  =1.0753

CH2FCOOH dissociates as:- CH2FCOOH  CH2FCOO-  + H+

For dissociation, = (i-1) /( n-1)

Here i=1.0753 and n=2

= 1.0753-1 = 0.0753

Therefore, Ka = [CH2FCOO-] [H+] / [CH2FCOOH]

= C2 /(1-)

= 0.50 x (0.0753)2 / 0.9247 =3.07 x 10-3

Related Questions
Sun July 17, 2016

# A solution X is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.8, then depression in freezing point of solution X is. ( Kf water = 1.86 and Kf ethanol = 2.0 ) 1. 4.34 K 2. 10.86 K 3. 5.434 K 4. 1.35 K

Sun July 10, 2016