5.
Specific volume of cylindrical virus particle is
6.02 × 10-2 cc/gm whose radius and length are
7 Å and 10 Å respectively. If Na = 6.02 x 1023,
find molecular weight of virus.
(1) 15.4 kg/mol
(2) 1.54 x 104 kg/mol
(3) 3.08 * 104 kg/mol (4) 3.08 x 103 kg/mol
Answers
Answered by
8
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Answered by
6
The correct option is : (a) 15.4 kg/mol
Explanation :
Volume of cylinder =πr
2
h
As the virus is cylindrical.
∴ Volume of one virus = volume of cylinder
As the radius and length of virus are 7
A
˚
and 10
A
˚
respectively.
∴ Volume of 1 virus =
7
22
×7
2
×10=1540×10
−24
cm
3
=1.54×10
−21
cm
3
[∵
1
˚
=10
−8
cm
3
]
Volume of 1 mole of virus =N
A
×volume of 1 virus
⇒ Volume of 1 mole of virus =6.02×10
23
×1.54×10
−21
=6.02×1.54×10
2
cm
3
Given that specific volume of virus is 6.02×10
−2
cm
3
/g
∴Molecular weight=
Specific volume
volume of 1 mole
=
6.02×10
−2
6.02×1.54×10
2
=1.54×10
4
g/mol=15.4Kg/mol
Hope this answer is helpful for u.
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