Answer:
mass(g) of colchicine = 3.99 × 10⁻⁵g
Explanation:
Given that;
the number of moles colchicine = 100 nM = 100 × 10⁻⁹M
Density of chlamydomonas culture = 5 × 10⁷ cells/mL
∴ In 1mL of chlamydomonas culture there are 5 × 10⁷ cells present.
To decide the number of one cell, we need to use [tex]\frac{1}{5*10^7}[/tex] mL of the cellulose
However, 1 × 10⁸ cells will be present in [tex]1*10^8*\frac{1}{5*10^7}[/tex] mL , which in turn give us;
[tex]\frac{10^8}{5*10^7}[/tex]mL
Afterwards, to get the required 1 × 10⁸ cells, 2mL(molarity, since molarity= no of moles/litre) has to be taken
If colchicine has to be treated, we need to determine the mass of colchicine that is required in the process as well;
since, the number of moles colchicine = 100 nM = 100 × 10⁻⁹M
And, the given molecular weight = 399.44;
we can determine the mass of colchicine as;
∴ [tex]numberofmoles = \frac{mass(g)}{molar mass(g)}[/tex]
substituting the parameters given, we have:
[tex]100*10^{-9}= \frac{mass(g)}{399.4}[/tex]
mass(g) = 399.4 × 100 × 10⁻⁹
= 3.99 × 10⁻⁵g
Hence, the mass of colchicine that is required in the process to make 100 nM dissolve in the in 2mL of the culture in one Litre of water is 3.99 × 10⁻⁵g.
