6. Specific volume of cylindrical virus particle is
6.02 x 10-2 cc/gm whose radius and length are
7 A and 10 A respectively. If NA = 6.02 * 1023,
find molecular weight of virus.
(1) 15.4 kg/mol (2) 1.54 * 104 kg/mol
(3) 3.08 * 104 kg/mol (4) 3.08 * 10 kg/mol
Answers
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Explanation:
Correct option is
A
1.54kg/mol
ℓ
1
=6.02×10
−2
cc/gm
r=7
A
˚
L=10
A
˚
V=πr
2
L
=π×(7×10
−8
)
2
(10×10
−8
)
M=
1/ℓ
V
=
6.02×10
2
π×49×10
−23
molecularweight=
6.02×10
2
49π×10
−23
×6.02×10
23
(
100
49π
)=1.54kg/mol
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