Question

specific volume of a cylindrical virus particle is 6.02× 10^-2 cc/gm whose radius and length are 7 angstrom and 10 angstrom respectively .if Avogadro number is 6.02 × 10^23. find molecular weight of virus

Answer 1

Hi...

volume of one virus = voume of cylinder = π r²h

= 22/7 × 7 × 7 × 10 A°³

=22 × 7 × 10 A°³

=1540 × 10^-30 m³

=1.54 × 10^-27 m³

so, volume of one mole of virus = 6.023 × 10²³ × volume of one virus

= 6.023 × 10²³ × 1.54 × 10^-27 m³

= 9.27542 × 10^-4 m³

= 9.27542 × 10² cm³

now, molecular weight = volume of 1mole/ specific volume

= 9.27542 ×10² / 6.02 × 10^-2 g/mol
= 1.54 × 10⁴ g/mol.....
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Hacker20 ✯ Brainly star ✯
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

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 particles

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

= 15400 gm/Mole



Hope this helps u!!