Question

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

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

Answer 1

Answer: (B)

Specific volume (volume of $1g$) of cylindrical virus particle
$=6.02\times 10^{-2}$ cc/g
Radius of virus (r) $=7\mathring{A}=7\times 10^{-8}cm$
Length of virus $=10\times 10^{-8}cm$
Volume of virus
$\pi r^2=\frac{22}{7}\times (7\times 10^{-8}{)}^{2}times10\times 10^{-8}$
$=154\times 10^{-23}cc$
Wt. of one virus particle $=\frac{\text{volume}}{\text{specific volume}}$
$\therefore$ Mol. wt. of virus $=$ wt. of $N_A$ particle
$=\frac{154\times 10^{-23}}{6.02\times 10^{-2}}\times 6.02\times 10^{23}$
$=15400g/mol=15.4kg/mol$