Question

Phosphorus-32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na₃³²PO₄ after 45.6 days? Assume that the atomic mass of ³²P is 32.0.

Answer 1

Using the half-life of 14.3 days, we calculate that after 45.6 days, approximately 12.24 mg of a 175 mg sample of Phosphorus-32 remains.

Step-by-step explanation:

The radioactive decay of phosphorus-32 (Phosphorus-32) can be calculated using its half-life information. Phosphorus-32 has a half-life of 14.3 days, which means that half of the isotope will decay every 14.3 days, resulting in half of the original mass remaining. To determine the mass of phosphorus-32 left after 45.6 days from an original sample of 175 mg, we first calculate the number of half-lives that have passed in 45.6 days. We do this by dividing 45.6 by the half-life period (14.3), giving us approximately 3.19 half-lives.

Using the half-life decay formula, we can calculate the remaining mass:

Remaining Mass = Initial Mass × (0.5)Number of Half-Lives

Remaining Mass = 175 mg × (0.5)3.19

Calculating this, we find that the remaining mass of Phosphorus-32 is approximately 12.24 mg.

Answer 2

The correct answer is 6.65 mg.

Step-by-step explanation:

Based on the given information, the number of days given is 45.6 days. The original mass of sodium phosphate give is 175 mg. The half-life of phosphorus-32 is 14.3 days, therefore, the n or the number of half life will be,

n = 45.6/14.3 = 3.19

So, after 45.6 days, the mass of sodium phosphate sample left will be,

= Original mass × 1/2ⁿ

= 175 mg × 1/2 ^3.19

= 175 mg × 0.1096

= 34.3 mg of sodium phosphate left after 45.6 days

The molecular mass of Na₃³²PO₄ is 3 (23) + 32 + 4 (16) amu = 165 amu

Therefore, 165 grams of sodium phosphate comprise 32 grams of phosphorus.

So, 34.3 mg or 0.343 gram of sodium phosphate will contain,

= 0.343 × 32/165

= 6.65 mg of P³².