Specific volume of cylindrical virus particle is $6.02 \times 10^{-2} \mathrm{cc} / \mathrm{g}$. whose radius and length are $7 Å \& 10 Å$ respectively.
If $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$, find molecular weight of virus

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Aicharya · Accepted Answer

Specific volume (volume of $1 \mathrm{~g}$ ) of cylindrical virus particle $=6.02 \times 10^{-2} \mathrm{cc} / \mathrm{g}$
Radius of virus $(\mathrm{r})=7 Å=7 \times 10^{-8} \mathrm{~cm}$
Length of virus $=10 \times 10^{-8} \mathrm{~cm}$
Volume of virus $=$
$\pi r^{2} 1=\frac{22}{7} \times\left(7 \times 10^{-8}ight)^{2} \times 10 \times 10^{-8}$
$=154 \times 10^{-23} \mathrm{cc}$
Wt. of one virus particle $=\frac{\text { volume }}{\text { specific volume }}$
$\therefore$ Mol. wt. of virus $=$ Wt. of $N_{A}$ particle
$=\frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \times 6.02 \times 10^{23}=15400 \mathrm{~g} / \mathrm{mol}$
$=15.4 \mathrm{~kg} / \mathrm{mol}$

Specific volume of cylindrical virus particle is $6.02 \times 10^{-2} \mathrm{cc} / \mathrm{g}$. whose…

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