In the spring of 1984, concern arose over the presence of ethylene dibromide, or EDB, in grains and cereals. EDB has the molecular formula C2H4Br2 and until 1984 was commonly used as a plant fumigant. The federal limit for EDB in finished cereal products is 30.0 parts per billion (ppb), where 1.0 ppb = 1.0 ✕ 10−9 g of EDB for every 1.0 g of sample. How many molecules of EDB are in 1.9 lb of flour if 31.5 ppb of EDB is present?

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cristisp93 cristisp93 · Answer 1 · 2019-10-26T17:00:16+02:00

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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jusmeannaa jusmeannaa · Answer 2 · 2020-05-08T17:24:50+02:00

Answer:

thank you

Explanation:

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