Answer:
The correct answer is 6.65 mg.
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.
