Question

A plant virus is found to consist of uniform cylindrical particles of $150 \,\mathring{A}$ in diameter and $5000 \,\mathring{A}$ long. The specific volume of the virus is $0.75 \,cm^3 / g$. If the virus is considered to be a single particle, find its molar mass.

Answer 1

Volume of one cylinderical plant virus $=\pi r^{2} l$
$=3.14\left(75 \times 10^{-8}\right)^{2} \times 5000 \times 10^{-8} cm ^{3}=8.83 \times 10^{-17} cm ^{3} $
$\Rightarrow $ Mass of one virus $=\frac{\text { Volume of a virus }}{\text { Specific volume }} $
$=\frac{8.83 \times 10^{-17} cm ^{3}}{0.75 cm ^{3} g ^{-1}}=1.1773 \times 10^{-16} g $
$\Rightarrow $ Molar mass of virus $=$ Mass of one virus $ \times $ Avogadro's number
$=1.1773 \times 10^{-16} \times 6.023 \times 10^{23} g $
$=70.91 \times 10^{6} g$