869.6 × 10¹⁴ molecules of EDB
Explanation:
We have 1.9 lb of flour with a EDB concentration of 31.5 ppb.
We need to transform lb in grams.
1 lb = 453.6 grams
1.9 lb = (1.9 × 453.6) / 1 = 861.8 grams
Now we determine the number of molecules of EDB in the sample by devise the following reasoning:
if we have 31.5 × 10⁻⁹ g of EDB in 1 g of sample
then we have X g of EDB in 861.8 g of sample
X = (31.5 × 10⁻⁹ × 861.8) / 1 = 27146.7 × 10⁻⁹ g of EDB
Molecular mass of EDB (C₂H₄Br₂) = 188 g/mole
Taking in account that 1 mole of any substance contains 6.022 × 10²³ (Avogadro’s number) molecules we devise the following reasoning:
if 188 g of EDB contains 6.022 × 10²³ molecules
then 27146.7 × 10⁻⁹ g of EDB contains Y molecules
Y = (27146.7 × 10⁻⁹ × 6.022 × 10²³) / 188 = 869.6 × 10¹⁴ molecules of EDB
Learn more:
about Avogadro’s number
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