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. `3.08 xx 10^(3) kg // mol`
B. `3.08 xx 10^(4) kg // mol`
C. `1.54 xx 10^(4) kg // mol`
D. `1.54 kg // mol`

Answer

D

Explanation

Specific volume (volume of 1 g) of cylindrical virus particle `=6.02 xx 10^(-2) c c // g`
Radius of virus (r) =7 Å `= 7 xx 10^(-8) cm`
Length of virus `= 10 xx 10^(-8) cm`
Volume of virus
`pi r^(2)l=(22)/(7) xx (7 xx 10^(-8) xx 10 xx 10^(-8))=154 xx 10^(23) c c`
Wt. of one virus particle `= ("volume")/("specific volume")`
 `:.` Mol. wt. of virus = Wt. of `N_(A)` particle
`= (154 xx 10^(-23))/(6.02 xx 10^(-2)) xx 6.02 xx 10^(23)= 15400 g // mol = 15.4 kg // "mole"`