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Specific volume of cylindrical virus par...

Specific volume of cylindrical virus particle is `6.02xx10^(-2)c c//g` , whose radius and length are `7Å and 10Å` respectively . If `N_(A)=6.023xx10^(23)` , find molecular weight of virus.

A

1.54 kg/mol.

B

`1.54 xx 10^4` kg/mol.

C

`3.08 xx 10^4` kg/mol.

D

`3.08 xx 10^3` kg/mol.

Text Solution

Verified by Experts

The correct Answer is:
A
Sp. Vol (vol. of 1gm ) cylindrical virus particle `=6.02xx10^(-2)` cc/gm
radius of virus r=7Å= `7xx10^(-8)` cm
length of virus =`pir^2l`
`=22/7xx(7xx10^(-8))^2xx10xx10^(-8)=154xx10^(-23)` cc
wt. of one virus particle =`"Vol."/"Sp. vol."`
`rArr (154xx10^(-23))/(6.02xx10^(-2))` gm
`therefore` mol. wt. of virus =wt. of `N_A` particles
`=(154xx10^(-23))/(6.02xx10^(-2))xx6.02xx10^(+23)` gm/mol
=15400 gm/mol =15.4 kg/mol
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