Question

Specific volume of cylindrical virus particle is $6.02\times {10}^{-2}\frac{cc}{g}$, whose radius and length arc $7\stackrel{\circ }{A}$ respectively. If $N_A=6.02\times {10}^{23},$ find molecular weight of virus.

Answer 1

Specific volume $=6.02\times {10}^{-\frac{2cc}{gm}}$
So density $=\frac{1}{specificvolume}=\frac{1}{6}.02\times {10}^{\frac{-2gm}{cc}}$
Volume of the cylinder $=\pi r^2\ast h=\pi \times 7\times 7\times 10\times 10-30m^3$
Hence, mass of one virus = density $\times$ volume
mass of avogadro number of virus = NA $\times$ mass of one virus = molecular weight of the virus.