Question

A plant virus is found to consist of uniform cylindrical
particle of 150 Angstrom in diameter and 5000 Angstrom long. The
specific volume of the virus is 0.75 mL /g. If the virus is
considered to be a single particle, find its molar mass​

Answer 1

Given :

Diameter = 150Å (Radius = 75Å)

Length of virus = 5000Å

Specific volume = 0.75mL/g

To Find :

Molar mass of the virus.

Solution :

❖ First of all we need to find volume of one cylindrical virus.

Volume of cylinder = π r² l

• r denotes radius
• l denotes length

» r = 75Å = 75 × 10‾⁸ cm

» l = 5000Å = 5000 × 10‾⁸ cm

By substituting the given values;

$\sf:\implies\:V=\pi\times(75\times10^{-8})^2\times(5000\times 10^{-8})$

$\sf:\implies\:V=3.14\times (5625\times10^{-16})\times(5\times10^{-5})$

$\sf:\implies\:V=88312.5\times10^{-21}$

$\bf:\implies\:V=8.83\times10^{-17}\:cm^3$

♦ Now let's calculate mass of one cylindrical virus.

Specific volume = 0.75 cm³/g

• we know that, 1 mL = 1 cm³

$\sf:\implies\:Mass=\dfrac{Volume}{Specific\:volume}$

$\sf:\implies\:Mass=\dfrac{8.83\times10^{-17}}{0.75}$

$\sf:\implies\:Mass=11.77\times 10^{-17}$

$\bf:\implies\:Mass=1.18\times 10^{-16}\:g$

♦ Molar mass of virus is given by

$\sf:\implies\:Molar\:mass=Mass_{one\:virus}\times N_A$

• $\sf{N_A}$ denotes avogadro's number

$\sf:\implies\:Molar\:mass=(1.18\times10^{-16})\times(6.022\times10^{23})$

$\sf:\implies\:\underline{\boxed{\bf{\orange{Molar\:mass=7.1\times10^{7}\:g\:mol^{-1}}}}}$

Answer 2

Explanation:

Percentage loss in mass=2×

266.5

18

×100=13.50%

Specific volume =

Mass

Volume

Mass=

Specificvolume

πr

2

h

=

0.75

3.14×(75×10

−8

)

2

×5000×10

−8

Molar mass=Mass×N =

0.5

3.14×(75×10

−8

)

2

×5000×10

−8

×6.023×10

23

=7.092×10

7

g