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. $15.4\, kg/mol$
• C. $1.54 \times 10^4\, kg/mol$
• D. $3.08 \times 10^4\, kg/mol$

Answer 1

Answer: (B)

Specific volume (volume of $1\, g$) 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
$\pi r^2 = \frac{22}{7} \times (7 \times 10^{-8} )^2 \ times 10 \times 10^{-8}$
$ = 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}$
$= 15400\, g/mol = 15.4 \,kg/mol$