Answer:
a) The volume is 5.236x10⁻¹³L
b) The molarity of a single virus is 1.91x10¹² mol/L
c) The molarity for a 100 virus particles is 1.91x10¹⁴ mol/L
Explanation:
a) Given:
D = diameter of the cell = 10 μm
r = radius = 10/2 = 5 μm
The volume of the spherical cell is equal:
[tex]V=\frac{4}{3} \pi r^{3} =\frac{4}{3} \pi *(5)^{3} =523.6\mu m^{3}[/tex]
If 1 μm³ = 1x10⁻¹⁵L, then 523.6 μm³ = 5.236x10⁻¹³L
b) The molarity is:
[tex]M=\frac{number-of-moles}{volume-of-solution}[/tex]
For a single virus within the cell
[tex]M=\frac{1}{5.236x10^{-13} } =1.91x10^{12} mol/L[/tex]
c) For a 100 virus particles the molarity is:
[tex]M=\frac{100}{5.236x10^{-13} } =1.91x10^{14} mol/L[/tex]
