The concentration of the alkane in a particular sample of rainwater is 0.2 ppb. assume that the density of rainwater is close to 1.00 g/ml and find the molar concentration of

Ppb stands for parts per billion 10⁻⁹.
M(C₂₀H₄₂, alkane) = 282,55 g/mol.
V(C₂₀H₄₂) = 1000 ml = 1 dm³.
m(C₂₀H₄₂) = 2·10⁻⁷ g.
n(C₂₀H₄₂) = m(C₂₀H₄₂) ÷ M(C₂₀H₄₂).
n(C₂₀H₄₂) = 2·10⁻⁷ g ÷ 282,55 g/mol.
n(C₂₀H₄₂) = 7,07·10⁻¹⁰ mol.
c(C₂₀H₄₂) = n(C₂₀H₄₂) ÷ V(C₂₀H₄₂).
c(C₂₀H₄₂) = 7,07·10⁻¹⁰ mol/dm³.

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