Réponse:

1)

N = n × Na

N = 20.10-³× 6,02.10²³

N = 1,2.10²⁵ molecules de dioxygène

2)

a.

m(C₁₃H₁₈O₂) = 13×1,99.10⁻²⁶ + 18×1,67.10⁻²⁷ + 2×2,66.10⁻²⁶

m(C₁₃H₁₈O₂) = 3,42.10⁻²⁵ kg

b.

N = 400.10⁻⁶/3,42.10⁻²⁵ ( les masses sont en kg)

N = 1,17.10²¹ molecules d'ibuprofene

c.

n = N/Na

n = 1,17.10²¹/6,02.10²³

n = 1,94.10-³ mol

3)

m(NaHCO₃) = 2,33.10⁻²⁶ + 1,67.10⁻²⁷ + 1,99.10⁻²⁶ + 3×2,66.10⁻²⁶

m(NaHCO₃) = 1,25.10⁻²⁵ kg

N(NaHCO₃) = 3.10-³/1,25.10⁻²⁵

N(NaHCO₃) = 2,40.10²² entités

n(NaHCO₃) = N(NaHCO₃) / Na

n(NaHCO₃) = 2,40.10²² / 6,02.10²³

n(NaHCO₃) = 3,99.10-² mol

4)

m(CH₄N₂O) = 1,99.10⁻²⁶ + 4×1,67.10⁻²⁷ + 2×2,33.10⁻²⁶ + 2,66.10⁻²⁶

m(CH₄N₂O) = 9,98.10⁻²⁶ kg

N(CH₄N₂O) = 0,49.10⁻³/ 9,98.10⁻²⁶

N(CH₄N₂O) = 4,91.10²¹ molecules

n(CH₄N₂O) = N(CH₄N₂O) /Na

n(CH₄N₂O) = 4,91.10²¹/6,02.10²³

n(CH₄N₂O) = 8,16.10⁻³ mol

5)

Supposons que l'échantillon soit de l'ozone :

m(O₃) = 3×2,66.10⁻²⁶

m(O₃) = 7,98.10⁻²⁶ kg

m(O₃) = 7,98.10⁻²³ g

N(O₃) = n × Na

N(O₃) = 1,0×6,02.10²³

N(O₃) = 6,02.10²³ molecules

m(O₃) × N(O₃) = 7,98.10⁻²³ × 6,02.10²³

m(O₃) × N(O₃) = 48,0 g

La masse d'une mole d'ozone est de 48,0 g

Facultatif :{

Supposons que l'échantillon soit du dioxygène

m(O₂) = 2×2,66.10⁻²⁶

m(O₂) = 5,32.10⁻²⁶ kg

m(O₂) = 5,32.10⁻²³ g

m(O₂) × N(O₂) = 32,0 g

la masse d'une mole de dioxygène est de 32,0 g

}

L'échantillon est donc de l'ozone.

6.

a.

n(C₇H₆O₃) = m1 / (Na × m(C₇H₆O₃) )

n(C₇H₆O₃) = 3,5/(6,02.10²³×2,29.10-²²)

n(C₇H₆O₃) = 2,5.10-² mol

b.

m₂ = ρ(C₄H₆O₃) × V₂

m₂ = 1,08×5,0

m₂ = 5,4 g

c.

n(C₄H₆O₃) = m₂ /(Na × m(C₄H₆O₃) )

n(C₄H₆O₃) = 5,4 /(6,02.10²³×1,69.10-²²)

n(C₄H₆O₃) = 5,3.10-² mol