Question

A plant virus is found to consist of uniform cylindrical particles of $150$ angstrom in diameter and $5000$ angstom long. The specific volume of virus is $0.75{cm}^3{g}^{-1}$. If the virus is considered to be a single particle, its molecular weight is :
(Given that volume of cylinder is ${\pi r}^{2}l$)

• A. $7.092\times {10}^7$
• B. $7.092\times {10}^6$
• C. $7.092\times {10}^5$
• D. $7.092\times {10}^9$

Answer 1

The correct option is A $7.092\times {10}^7$

Volume of 1 cylindrical particle $={\pi r}^{2}l=3.1416\times (75\times {10}^{-8}{)}^2\times 5000\times {10}^{-8}=8.84\times {10}^{-17}$ $c{m}^3$.
Volume of Avogadro's number of particles $=8.84\times {10}^{-17}\times 6.023\times {10}^{23}=5.32\times {10}^7$ $c{m}^3$.
Molecular weight of particle $=\frac{5.32\times {10}^7}{0.75}=7.092\times {10}^7$.
This is equal to the molecular weight of virus.