The activity of a freshly prepared radioactive sample of a...

The activity of a radioactive sample is given by $A = \lambda N$, where $\lambda$ is the decay constant and $N$ is the number of active nuclei.

The average life $\tau$ is related to the decay constant by $\tau = \dfrac{1}{\lambda}$.

Substituting $\lambda = \dfrac{1}{\tau}$ into the activity equation, we get:

$A = \dfrac{N}{\tau} \Rightarrow N = A \tau$

Given $A = 10^{10}\ \texts^{-1}$ and $\tau = 10^9\ \text{s}$, the number of atoms is:

$N = 10^{10} \times 10^9 = 10^{19}$

The total mass $M$ of the radioactive sample is the product of the number of atoms and the mass of a single atom $m$:

$M = N \times m$

Given $m = 10^{-25}\ \text{kg}$, we have:

$M = 10^{19} \times 10^{-25}\ \text{kg} = 10^{-6}\ \text{kg}$

Converting the mass into milligrams ($1\ \text{kg} = 10^6\ \text{mg}$):

$M = 10^{-6} \times 10^6\ \text{mg} = 1\ \text{mg}$

Answer: $1\ \text{mg}$