Sp. vol. of cylindrical virus particle is $6.02 \times 10^{-2}$ cc/gm, whose radius and length are 7 Å and 10 Å respectively. If $N_A = 6.02 \times 10^{23}$, find mol. wt. of virus : -
A
15.4 kg/mol
B
$1.54 \times 10^4$ kg/mol
C
$3.08 \times 10^4$ kg/mol
D
$3.08 \times 10^3$ kg/mol
The answer is A.
Solution
Specific volume = volume of 1 g $= 6.02 \times 10^{-2}$ cc/g. Radius $r = 7$ Å $= 7 \times 10^{-8}$ cm; length $l = 10$ Å $= 10 \times 10^{-8}$ cm Volume of one virus $= \pi r^2l = \dfrac{22}{7} \times (7 \times 10^{-8})^2 \times 10 \times 10^{-8} = 154 \times 10^{-23}$ cc Mass of one virus particle $= \dfrac{\text{volume}}{\text{sp. volume}} = \dfrac{154 \times 10^{-23}}{6.02 \times 10^{-2}}$ g $\therefore$ mol. wt. = mass of $N_A$ particles $= \dfrac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \times 6.02 \times 10^{23}$ g/mol $= 15400$ g/mol $= 15.4$ kg/mol
ConceptMolar mass of a virus particle
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