Specific volume of cylindrical virus particle is 6.02 * 10-2 cc/gm whose radius and length are 7angstrom and 10angstrom respectively.If NA=6.02 *10^23 find molecular weight of virus???

Given data:

Specific volume (vol. of 1 gm) of cylindrical virus particle = 6.02x10^-2 c.c/gm

Radius of the virus (r) = 7A = 7x10^-8 cm

Length of the virus (l) = 10A = 10x10^-8 cm

NA (Avagadro’s number) = 6.023x10^23

Volume of virus = πr^2l

  = 3.14*(7x10^-8)^2*10x10^-8 cm

  = 154x10^-23 c.c

So, weight of 1 virus particle = Volume / Specific volume

  = 154x10^-23/6.02x10^-2 gm

Molecular weight of virus = Weight of NA particles

  = (154x10^-23/6.02x10^-2)* 6.023x10^23

  = 15400 gm/mole

  = 15.4 kg/mole