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Some Basic Concepts of Chemistry
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13/08/2021 11:03 am
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Specific volume of cylindrical virus particle is 6.02 × 10-2 cc/g. whose radius and length are 7 Å & 10 Å respectively.
If NA = 6.02 × 1023 mol1, find molecular weight of virus
(a) 3.08 × 103 kg/mol
(b) 3.08 × 104 kg/mol
(c) 1.54 × 104 kg/mol
(d) 15.4 kg/mol
1 Answer
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13/08/2021 11:11 am
Correct answer: (d) 15.4 kg/mol
Explanation:
Specific volume (volume of 1 g) of cylindrical virus particle = 6.02 × 10-2 cc/g
Radius of virus (r) = 7 Å = 7 × 10-8 cm
Length of virus = 10 × 10-8 cm
Volume of virus
= \(\pi r^2 I = \frac{22}{7}\) x (7 x 10-8)2 x 10 x 10-8
= 154 x 10-23cc
Wt of. one virus particle = \(\frac{volume}{specific\;volume}\)
∴ Mol. wt. of virus = Wt. of virus = Wt. of NA particle
= \(\frac{154 \times 10^{-23}}{6.02 \times 10^{-2}}\) x 6.02 x 1023
= 15400 g/mol
= 15.4 kg/mol