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# Sp. vol. of cylindrical virus particle is 6.02 x 10^-2 cc/gm whose radius and length are 7Å and 10Å respectively. Find the molar wt. of the virus.

Naveen Kumar
7 years ago
Specific volume=6.02*10^-2cc/gm
So density=1/specific volume=1/6.02*10^-2 gm/cc
volume of the cylinder=π*r2*h=π*7*7*10*10-30 m3
Hence Mass of one Virus= density*Volume
Mass of avogadro number of virus=NA* mass of one virus=molecular weight of the virus.

Dheeraj jareda
20 Points
4 years ago
Specific volume (vol. of 1 gm) of cylindrical virus particle = 6.02x10-2 c.c/gRadius of the virus (r) = 7A = 7 × 10-8 cmLength of the virus (l) = 10A = 10 × 10-8 cmNA (Avagadro’s number) = 6.023 × 1023Volume 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 gMolecular 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