Respuesta :

Answer is: 6.022·10²² molecules of glucose.

c(glucose) = 100 mM.

c(glucose) = 100 · 10⁻³ mol/L.

c(glucose) = 0.1 mol/L; concentration of glucose solution.

V(glucose) = 1 L; volume of glucose solution.

n(glucose) = c(glucose) · V(glucose).

n(glucose) = 0.1 mol/L · 1 L.

n(glucose) = 0.1 mol; amount of substance.

N(glucose) = n(glucose) · Na (Avogadro constant).

N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.

N(glucose) = 6.022·10²².

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