Question

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

• A. $15.4kg/mol$
• B. $1.54\times 10^{4}kg/mol$
• C. $3.08\times 10^{4}kg/mol$
• D. $3.08\times 10^{3}kg/mol$

Answer 1

Answer: (A)

One molecule $(gm)$
$d=\frac{\text{ M. wt. }}{V}$
$\frac{1}{6.02\times 10^{-2}}=\frac{\text{ M. wt. }}{\pi r^2\times h}$
$M.Wt$. (One molecule in gm)
$=\frac{\pi \times {\left(7\times 10^{-8}\right)}^2\times 10\times 10^{-8}}{6.02\times 10^{-2}}$
$M.Wt$. (One mole in $kg$ )
$=\frac{22}{7}\times \frac{7\times 7\times 6.02\times 10^{-3}}{6.02\times 10^{-2}}$
$=2.2\times 7=15.4kgmo{l}^{-1}$