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 $0.75c{m}^3/g$. If the virus is considered to be a single particle, find its molar mass.

Answer 1

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