Chapter 9, Problem 9.48AP

(a)

Interpretation Introduction

Interpretation:

The amount of Phosphorous-32 and Sulfur-32 remains after $14\text{d}$ has to be given.

Answer to Problem 9.48AP

The amount of Phosphorous-32 remains after $14\text{d}$ is $62\text{mg}\text{.}$

The amount of Sulfur-32 remains after $14\text{d}$ is $62\text{mg}\text{.}$

Explanation of Solution

Given,

The mass of Phosphorous-32 at initial is $\text{124}\text{mg}\text{.}$

The half-life of Phosphorous-32 is $14\text{d}\text{.}$

The time interval is $14\text{d}\text{.}$

The number of days can be converted to the number of half life by,

  $\text{14}\text{d×}\frac{\text{1}\text{half}\text{life}}{\text{14}\text{d}}\text{=1}\text{Half}\text{Life}$

The amount of substance remains after half lives can be calculated by multiplying half lives with the initial mass,

The amount of substance remains after one half-lives (m) $\text{=124}\text{mg×}\frac{\text{1}}{\text{2}}$

  $\text{m=}\frac{124\text{mg}}{2}$

  $\text{m=}62\text{mg}$

The amount of Phosphorous-32 remains after $14\text{d}$ is $62\text{mg}\text{.}$

The amount of Sulfur-32 after $14\text{d}$ can be calculated as,

The amount of Sulfur-32=$\left(\text{Total}\text{mass-Mass}\text{of}\text{iodine}\right)$

The amount of Sulfur-32=$\left(\text{124-62mg}\right)$

The amount of Sulfur-32=$\text{62}\text{mg}$

The amount of Sulfur-32 remains after $14\text{d}$ is $62\text{mg}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The amount of Phosphorous-32 and Sulfur-32 remains after $28\text{d}$ has to be given.

Answer to Problem 9.48AP

The amount of Phosphorous-32 remains after $28\text{d}$ is $31\text{mg}\text{.}$

The amount of Sulfur-32 remains after $28\text{d}$ is $93\text{mg}\text{.}$

Explanation of Solution

Given,

The mass of Phosphorous-32 at initial is $\text{124}\text{mg}\text{.}$

The half-life of Phosphorous-32 is $14\text{d}\text{.}$

The time interval is $28\text{d}\text{.}$

The number of days can be converted to the number of half life by,

  $28\text{d×}\frac{\text{1}\text{half}\text{life}}{\text{14}\text{d}}\text{=2}\text{Half}\text{Life}$

The amount of substance remains after half lives can be calculated by multiplying half lives with the initial mass,

The amount of substance remains after two half-lives (m) $\text{=124}\text{mg×}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{\text{2}}$

  $\text{m=}\frac{124\text{mg}}{4}$

  $\text{m=}31\text{mg}$

The amount of Phosphorous-32 remains after $28\text{d}$ is $31\text{mg}\text{.}$

The amount of Sulfur-32 after $28\text{d}$ can be calculated as,

The amount of Sulfur-32=$\left(\text{Total}\text{mass-Mass}\text{of}\text{iodine}\right)$

The amount of Sulfur-32=$\left(\text{124-31mg}\right)$

The amount of Sulfur-32=$93\text{mg}$

The amount of Sulfur-32 remains after $28\text{d}$ is $93\text{mg}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The amount of Phosphorous-32 and Sulfur-32 remains after $42\text{d}$ has to be given.

Answer to Problem 9.48AP

The amount of Phosphorous-32 remains after $42\text{d}$ is $41.3\text{mg}\text{.}$

The amount of Sulfur-32 remains after $42\text{d}$ is $\text{82}\text{.7}\text{mg}\text{.}$

Explanation of Solution

Given,

The mass of Phosphorous-32 at initial is $\text{124}\text{mg}\text{.}$

The half-life of Phosphorous-32 is $14\text{d}\text{.}$

The time interval is $42\text{d}\text{.}$

The number of days can be converted to the number of half life by,

  $42\text{d×}\frac{\text{1}\text{half}\text{life}}{\text{14}\text{d}}\text{=3}\text{Half}\text{Life}$

The amount of substance remains after half lives can be calculated by multiplying half lives with the initial mass,

The amount of substance remains after three half-lives (m) $\text{=124}\text{mg×}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{\text{2}}$

  $\text{m=}\frac{124\text{mg}}{3}$

  $\text{m=}41.3\text{mg}$

The amount of Phosphorous-32 remains after $42\text{d}$ is $41.3\text{mg}\text{.}$

The amount of Sulfur-32 after $42\text{d}$ can be calculated as,

The amount of Sulfur-32=$\left(\text{Total}\text{mass-Mass}\text{of}\text{iodine}\right)$

The amount of Sulfur-32=$\left(\text{124-41}\text{.3mg}\right)$

The amount of Sulfur-32=$82.7\text{mg}$

The amount of Sulfur-32 remains after $42\text{d}$ is $\text{82}\text{.7}\text{mg}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The amount of Phosphorous-32 and Sulfur-32 remains after $56\text{d}$ has to be given.

Answer to Problem 9.48AP

The amount of Phosphorous-32 remains after $56\text{d}$ is $7.75\text{mg}\text{.}$

The amount of Sulfur-32 remains after $56\text{d}$ is $116.25\text{mg}\text{.}$

Explanation of Solution

Given,

The mass of Phosphorous-32 at initial is $\text{124}\text{mg}\text{.}$

The half-life of Phosphorous-32 is $14\text{d}\text{.}$

The time interval is $56\text{d}\text{.}$

The number of days can be converted to the number of half life by,

  $56\text{d×}\frac{\text{1}\text{half}\text{life}}{\text{14}\text{d}}\text{=4}\text{Half}\text{Life}$

The amount of substance remains after half lives can be calculated by multiplying half lives with the initial mass,

The amount of substance remains after one-half lives (m) $\text{=124}\text{mg×}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{\text{2}}$

  $\text{m=}\frac{124\text{mg}}{16}$

  $\text{m=7}\text{.75}\text{mg}$

The amount of Phosphorous-32 remains after $56\text{d}$ is $7.75\text{mg}\text{.}$

The amount of Sulfur-32 after $56\text{d}$ can be calculated as,

The amount of Sulfur-32=$\left(\text{Total}\text{mass-Mass}\text{of}\text{iodine}\right)$

The amount of Sulfur-32=$\left(\text{124-}31\text{mg}\right)$

The amount of Sulfur-32=$116.25\text{mg}$

The amount of Sulfur-32 remains after $56\text{d}$ is $116.25\text{mg}\text{.}$