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. $15.4 kg/mol$
• 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)

$15.4 kg/mol$

Answer 2

Specific volume (volume of 1g) cylindrical virus particle
$= 6.02 \times 10^{-2} cc/g$
Radius of virus (r) $= 7 \times 10^{-8} cm$
Length of virus (I) $= 10 \times 10^{-8} cm$
$= 154 \times 10^{-23} cc$
Weight of one virus particle
$\frac {Volume }{Specific \, volume} = \frac {154 \times 10^{23}}{6.02 \times 10^{-2}}$
$\therefore$ Molecular weight of vims = weight of $N_A$ particles
$= \frac {154 \times 10^{-23}}{6.02 \times 10^{-2}} \times 6.023 \times 10^{23}$
= 15400 g/mol = 15.4 kg/mol