Question

Question 1.

Answer 1

Answer:

Question 1.  → The amount remained is 15.625 g after 5 half-lives (30 hours).

Question 2. →  mass % of water =41.4 %

Question 3. →  C. 64 grams.

Explanation:

Question 1.

It is known that the decay of isotopes and radioactive material obeys first order kinetics.

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

Thus, 30 hours represent (30/6) 5 half-lives.

500 g →(first 1/2 life) 250 g →(second 1/2 life) 125 g →(third 1/2 life) 62.5 g →(fourth 1/2 life) 31.25 g →(fifth 1/2 life) 15.625 g

So, The amount remained is 15.625 g after 5 half-lives (30 hours).

Question 2.

• The formula of Copper (II) Fluoride tetra hydrate is CuF₂.4H₂O.

• the molar mass of Copper (II) Fluoride tetra hydrate is the sum of the atomic masses o each element.

• molar mass of 1 mol of CuF₂.4H₂O = 64 + 2*(19) + 8* (1) + 4*(16) = 174 g/mol.

• molar mass of 4 mol of H₂O = 8*(1) + 4*(16) = 72 g/mol

• 1 mol of Copper (II) Fluoride tetra hydrate → 4 mol H₂O

mass % of water = (mass of 4 mol of H₂O) / (mass of 1 mol of CuF₂.4H₂O ) *100 = (72 /174) *100 = 41.4 %

Question 3.

Mass percentage is defined as the mass of a solute divided by the total mass of the solution, multiplied by 100

• mass percent = ( mass of solute / mass of solution) *100

∴ 16 = ( mass of solute / 400 g) *100

mass of solute = (16* 400)/ 100 = 64 g.

So the right choice is C. 64 grams.