Question

9. Specific volume of cylindrical virus particle
is 6.02 x 10-2 ccig, whose radius and length
are 78 and 108 respectively. If
NA = 6.02310, find molecular weight of
virus.
(2001)
197 15.4 kg/mol (b) 1.54 x 104 kg/mol
(C) 3.08 x 104 kg/mol (0) 3.08 x 103 kg/mol​

Answer 1

Answer:

(B) option is correct

Explanation:

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