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. `15.4 kg//mol`
B. `1.54xx10^(4)kg//mol`
C. `3.08xx10^(4)kg//mol`
D. `3.08xx10^(3)kg//mol`

Answer

A

Explanation

Specific volume (volume of 1 g )cylindrical virus particle
`=6.02xx10^(-2) c c//g`
Radius of virus ( r )`=7xx10^(-8)cm`
Length of virus (1) `=10xx10^(-8)cm`
Volume of virus
`=pir^(2)1=(22)/(7)xx(7xx10^(-8))^(2)xx10xx10^(-8)`
`=154xx10^(-23)c c`
weight of one virus particle
`=("Volume")/("Specific volume")=(154xx10^(-23))/(6.02xx10^(-2))`
`:.` Molecular weight of virus = weight of `N_(A)` particles
`=(154xx10^(-23))/(6.02xx10^(-2))xx6.023xx10^(23)`
`=15400g//mol=15.4 kg//mol`