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

Options

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.

Answer

A

Explanation

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