Chapter 29, Problem 51AP

(a)

To determine

The number of moles of $\text{C}$.

Answer to Problem 51AP

The number of moles of $\text{C}$ is $3.18\times {\text{10}}^{-7}\text{mol}$.

Explanation of Solution

Given Info: The sample contains $3.5\text{μg}$ of $\text{C}$. Half-life of $\text{C}$ is 20.4 min.

Formula to calculate the number of moles of $\text{C}$ is,

$n=\frac{m}{m_C{N}_A}$

• m is the mass of the sample.
• $m_C$ is the mass of $\text{C}$.
• $N_A$ is the Avogadro number.

Substitute $3.5\text{μg}$ for m, 11.011 u for $m_C$ and $6.022\times {10}^{23}\text{atoms/mol}$ for $N_A$.

$\begin{array}{c}n=\frac{3.5\text{μg}}{\left(11.011\text{u}\right)\left(6.022\times {10}^{23}\text{atoms/mol}\right)}\\ =\frac{3.5\times {\text{10}}^{-6}\text{g}}{\left(11.011\text{u}\right)\left(1.661\times {10}^{-27}\text{kg/u}\right)\left(6.022\times {10}^{23}\text{atoms/mol}\right)}\\ =3.18\times {\text{10}}^{-7}\text{mol}\end{array}$

Conclusion:

The number of moles of $\text{C}$ is $3.18\times {\text{10}}^{-7}\text{mol}$.

(b)

To determine

The initial number of nuclei.

Answer to Problem 51AP

The initial number of nuclei is $1.91\times {10}^{17}$.

Explanation of Solution

Given Info: The sample contains $3.5\text{μg}$ of $\text{C}$. Half-life of $\text{C}$ is 20.4 min.

Formula to calculate the initial number of nuclei is,

$N_0=n{N}_A$

Substitute $3.18\times {\text{10}}^{-7}\text{mol}$ for n and $6.022\times {10}^{23}\text{atoms/mol}$ for $N_A$.

$\begin{array}{c}{N}_0=\left(3.18\times {\text{10}}^{-7}\text{mol}\right)\left(6.022\times {10}^{23}\text{atoms/mol}\right)\\ =1.91\times {10}^{17}\end{array}$

Conclusion:

The initial number of nuclei is $1.91\times {10}^{17}$.

(c)

To determine

The initial activity.

Answer to Problem 51AP

The initial activity is $1.08\times {10}^{14}\text{Bq}$.

Explanation of Solution

Given Info: The sample contains $3.5\text{μg}$ of $\text{C}$. Half-life of $\text{C}$ is 20.4 min.

Formula to calculate the initial activity is,

$R_0=\frac{N_0\ln 2}{t_{1/2}}$

• $t_{1/2}$ is the half-life.

Substitute $1.91\times {10}^{17}$ for $N_0$ and 20.4 min for $t_{1/2}$

$\begin{array}{c}{R}_0=\frac{\left(1.91\times {10}^{17}\right)\ln 2}{20.4\min }\\ =\frac{\left(1.91\times {10}^{17}\right)\ln 2}{\left(20.4\min \right)\left(\frac{\text{60}\text{s}}{1\min }\right)}\\ =1.08\times {10}^{14}\text{Bq}\end{array}$

Conclusion:

The initial activity is $1.08\times {10}^{14}\text{Bq}$.

(d)

To determine

The activity after 8.0 h.

Answer to Problem 51AP

The activity after 8.0 h is $8.92\times {10}^6\text{Bq}$.

Explanation of Solution

Given Info: The sample contains $3.5\text{μg}$ of $\text{C}$. Half-life of $\text{C}$ is 20.4 min.

Formula to calculate the activity after 8.0 h is,

$R=R_0\exp \left[\left(\frac{t}{t_{1/2}}\right)\ln 2\right]$

• $t_{1/2}$ is the half-life.

Substitute $1.08\times {10}^{14}\text{Bq}$ for $R_0$, 8.0 h for t and 20.4 min for $t_{1/2}$

$\begin{array}{l}R=\left(1.08\times {10}^{14}\text{Bq}\right)\exp \left[\left(\frac{8.0\text{h}}{20.4\min }\right)\ln 2\right]\\ =\left(1.08\times {10}^{14}\text{Bq}\right)\exp \left[\left(\frac{8.0\text{h}}{\left(20.4\min \right)\left(\frac{1\text{h}}{60\min }\right)}\right)\ln 2\right]\\ =8.92\times {10}^6\text{Bq}\end{array}$

Conclusion:

The activity after 8.0 h is $8.92\times {10}^6\text{Bq}$