Specific gravity of cylindrical virus particle is 6.02 x 10 ^{-2} cc/gm whose radius and length are 7 angstrom and 10 angstrom respectively . If Na is 6.022 x 10^{23}, find molecular wt of virus. Ma’am please explain the meaning of the question as well as the answer.

**Data:**

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

Radius of the virus r = 7A = 7 × 10^{-8}cm

Length of the virus (l) = 10A = 10 × 10^{-8}cm

NA (Avagadro’s number) = 6.023 × 10^{23}

Volume of virus = πr2l

= 3.14 × (7x10^{-8})2 × 10 × 10^{-8}cm

= 154 × 10^{-23} c.c.

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

= 154 × 10^{-23} /6.02 × 10^{-2} g

Molecular weight of virus = Weight of Sodium (Na) particles

= (154 × 10^{-23}/6.02 × 10^{-2} ) × 6.023 × 10^{23}

= 15400 gm/mole

= 15.4 kg/mole