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 1023, 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 × 10-8cm
Length of the virus (l) = 10A = 10 × 10-8cm
NA (Avagadro’s number) = 6.023 × 1023
Volume of virus = πr2l
= 3.14 × (7x10-8)2 × 10 × 10-8cm
= 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 × 1023
= 15400 gm/mole
= 15.4 kg/mole