Question

If the volume of acetic acid is 10 ml and its density is 0.6 g cm ,then the no of hydrogen atom are​

Answer 1

Explanation:

number of moles of CH

3

COOH=

(60)g/mol

(0.6)ml×(1.06)g/ml

=0.0106mol=n

Molarity=

1000ml×1gmol

−1

0.0106mol

=0.0106

Kg

mol

ΔT

f

=(1.86KKgmol

−1

)(0.0106molKg

−1

)=0.0197K

Von't Hoff Factor=

Calculated

Observed

FreezingPoint=

0.0197

0.0205

=1.041

CH

3

COOH⇋H

+

+CH

3

COO

−

n 0 0

n(1−x) nx nx

total moles=n−nx+2nx=n−nx

i=

n

n(1+x)

=1+x=1.041

Degree of Dissociation; x=0.041

[CH

3

COOH]=n(1−x)=(0.0106)(1−0.041)

[CH

3

COO

−

]=[H

+

]=nx=(0.0106)(0.041)

K

a

=

[CH

3

COOH]

[CH

3

COO

−

][H

+

]

=

(0.0106)(1−0.041)

(0.0106)

2

(0.041)

2

=1.86×10

−5