Specific volume of cylindrical virus particle is 6.02 into 10 to the power minus 2 cc per gm whose radius and length are 7 angstrom and then angstrom respectively find molecular weight of virus
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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
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
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