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Grade: 12th pass

                        

specific volume of cylindrical virus particle 6.02/100cc/gm whose radiud and length are7 angstrom and 10angstrom respectively find mol.wt. of virus

one year ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
10233 Points
							Dear student

Expression for specific volume is incorrect or unclear,

However it is reciprocal of density

Volume times density will give mass of the cylindrical virus, hope this helps

Regards
one year ago
Arun
25333 Points
							
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
 
one year ago
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