Specific volume of cylindrical virus particle is 6.02 × 10−2   cc/g whose radius and length 7  A0  and  10  A0 respectively. If NA=6.02 × 1023, find molecular weight of virus

Specific volume (volume of g) of cylindrical virus particle =6.02 × 10−2   cc/gRadius of virus (r) = 7  A0  = 7 × 10−8   cmLength  of  virus  =  10 × 10−8   cmVolume  of  virusπr2 L= 227 × 7 × 10−8  2 × 10 × 10−8  =154 × 10−23  ccWt.  of  one  virus  particle  = volumespecific  volume∴   Mol.  wt.  of virus =  wt.  of NA particle= 154 × 10−23   6.02 × 10−2    × 6.02 × 1023  =15400  g/mol=15.4  kg/mol