Question

Specific volume of cylindrical virus particle is $6.02\times 10^{-2}cc/g$, whose radius and length are $7\mathring{A}$ and $10\mathring{A}$ respectively. If $N_A=6.023\times 10^{23}$, find molecular weight of virus.

• A. $1.54kg/mol$
• 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)

$\frac{1}{\ell }=6.02\times 10^{-2}cc/gm$
$r=7\mathring{A}L=10\mathring{A}$
$V=\pi r^{2}L$
$=\pi \times {\left(7\times 10^{-8}\right)}^2\left(10\times 10^{-8}\right)$
$M=\frac{V}{1/\ell }=\frac{\pi \times 49\times 10^{-23}}{6.02\times 10^2}$
$\text{ molecular weight }=\frac{49\pi \times 10^{-23}\times 6.02\times 10^{23}}{6.02\times 10^2}$
$\left(\frac{49\pi }{100}\right)=1.54kg/mol$