Question

Specific volume of cylindrical virus particle is 6.02 × 10⁻² cc/gm.
whose radius and length 7 Å & 10 Å respectively.
If NA = 6.02 × 10²³, find molecular weight of virus
(a) 3.08 × 10³ kg/mol (b) 3.08 × 10⁴ kg/mol
(c) 1.54 × 10⁴ kg/mol (d) 15.4 kg/mol

Answer 1

Therefore the molecular weight of virus =4.25×10⁻⁴⁴ mole/gram.

Explanation:

Given, specific volume of cylinder virus particle is 6.02×10⁻² cc/gm

It means the volume of 1 gram of virus is 6.02×10⁻² cc

                                                                   =6.02×10⁻² cm³

                                                                   =6.02×10⁻² ×(×10⁻²)³ m³

                                                                   =6.02×10⁻⁸ m³

The radius of the virus = 7$\AA$=7×10⁻¹⁰m

The length of the virus = 10$\AA$ =10×10⁻¹⁰m

Therefore the volume of the virus=$\pi r^2h$

                                                      $=\pi (7\times 10^{-10})^2\times (10 \times 10^{-10})$ m³

                                                     =154×10⁻²⁹  m³

The total weight of the virus

$W_{total} =\frac{154 \times 10^{-29}}{6.02\times10^{-8}}$

         $=25.58\times 10^{-21}$ gm

Given, NA = 6.023×10²³

Therefore the molecular weight of virus$=\frac{W_{total}}{NA}$

                                                                   $=\frac{25.58\times 10^{-21}}{6.023\times 10^{23}}$  mole/gram

                                                                  =4.25×10⁻⁴⁴ mole/gram.