Some Basic Concepts of Chemistry - Result Question 19
####19. Specific volume of cylindrical virus particle is $6.02 \times 10^{-2} cc / g$ whose radius and length are $7 \AA$ $& 10 \AA$ respectively. If $N_A=6.02 \times 10^{23}$, find molecular weight of virus
[2001]
(a) $3.08 \times 10^{3} kg / mol$
(b) $3.08 \times 10^{4} kg / mol$
(c) $1.54 \times 10^{4} kg / mol$
(d) $15.4 kg / mol$
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Solution:
- (d) Specific volume (volume of $1 g$ ) of cylindrical virus particle $=6.02 \times 10^{-2} cc / g$
Radius of virus $(r)=7 \AA=7 \times 10^{-8} cm$
Length of virus $=10 \times 10^{-8} cm$
Volume of virus $=$
$\pi r^{2} l=\frac{22}{7} \times(7 \times 10^{-8})^{2} \times 10 \times 10^{-8}$
$=154 \times 10^{-23} cc$
Wt. of one virus particle $=\frac{\text{ volume }}{\text{ specific volume }}$
Mol. wt. of virus $=$ Wt. of $N_A$ particle
$=\frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \times 6.02 \times 10^{23}$
$=15400 g / mol=15.4 kg / mol$
