Question

19. Specific volume of cylindrical virus particle
is 6.02 x 10-2 cc'g, whose radius and length
are 78 and 108 respectively. If
No = 6.023 x 1023, find molecular weight of
virus.
(2001)
(a) 5.4 kg/mol (b) 1.54 x 104 kg/moi
(c) 3.08 x 104 kg/mol (0) 3.08 x 109 kg/mol
​

Answer 1

HX →H

+

+X

−

weak acid          0         0

1-a                       a         a

a = 20% dissociation

i=(1−a)+a+a=1+a=1+0.2=1.2

ΔT

f

​

=i×K

f

​

×m

=1.2×1.86×0.5

=1.12K