The number of mole in 27 μg of tetrahydrocannabinol (THC), C₂₁H₃₀O₂ is 8.6×10¯⁸ mole
We'll begin by converting 27 μg to grams (g). This can be obtained as follow:
1 μg = 1×10¯⁶ g
Therefore,
27 μg = 27 × 1×10¯⁶
Thus, 27 μg is equivalent to 27×10¯⁶ g.
Next, we shall determine the molar mass of C₂₁H₃₀O₂. This is illustrated below:
Molar mass of C₂₁H₃₀O₂ = (12×21) + (1×30) + (16×2)
= 252 + 30 + 32
Finally, we shall determine the number of mole in the 27 μg (i.e 27×10¯⁶ g) of C₂₁H₃₀O₂.
Molar mass of C₂₁H₃₀O₂ = 314 g/mol
Mass of C₂₁H₃₀O₂ = 27×10¯⁶ g.
[tex]Mole = \frac{mole}{molar mass}\\\\[/tex]
Mole of C₂₁H₃₀O₂ = [tex]\frac{27*10^{-6} }{314}\\\\[/tex]
Therefore, the number of mole in 27 μg (i.e 27×10¯⁶ g) of C₂₁H₃₀O₂ is 8.6×10¯⁸ mole
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