Question

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

• A. $1.54 \,kg/mol$
• B. $1.54 \times10^4\, kg/mol$
• C. $3.08 \times 10^4\,kg/mol $
• D. $3.08 \times10^3 \,kg/mol$

Answer 1

Answer: (A)

$\frac{1}{\ell}=6.02 \times 10^{-2} cc / gm $
$r =7 \mathring{A} \quad 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} $
$\left(\frac{49 \pi}{100}\right)=1.54 kg / mol$