2.8 × 10⁻³ mol of CO₂ is left after removing 10²¹ molecules from its ' x ' mg sample. The mass of CO₂ taken initially is Given: N_A=6.02 × 10²³ mol⁻¹

Solution

Chemistry ma’am · 3 steps

Turn the removed molecules into moles, add them back to what is left, then convert to milligrams.

1. Moles removed. Divide the number of molecules by Avogadro's number: 10²¹/(6.02 × 10²³) = 1.66 × 10⁻³ mol.
2. Moles at the start. The starting sample is what is left plus what was taken out: 2.8 × 10⁻³ + 1.66 × 10⁻³ = 4.46 × 10⁻³ mol.

   m = n × M = 4.46 × 10⁻³ mol × 44 g mol⁻¹ = 0.1962 g

3. So. 0.1962 g = 196.2 mg. The answer is option (b).

The idea to keep. Removing or adding molecules is just moles in disguise. Bring every quantity to moles first, do the bookkeeping there, and convert to mass only at the end.

Answer from the NTA final answer key. Solution by PrepWiser.