# Sodium-24 has ahalf-life of about 15 hours how much of an 1.60 gram sample of sodium-24 will remain after 60.0 hours? Ans :(1g)?

Apr 10, 2018

$0.1 g$

#### Explanation:

As we start with 1.6 grams and we know that 60 hours have passed, this equals 4 half-lives. This means that the weight of the sample has effectively been halved 4 times:

${\left(0.5\right)}^{4}$

So simply multiply this with the mass:

${\left(0.5\right)}^{4} \times 1.6 = 0.1 g$

Generally:
Remaining mass$= {\left(0.5\right)}^{n} \times m$

Where $n$ is the number of half-lives passed and $m$ is the original mass of the sample.

Apr 10, 2018

There seems to be a mistake in your given answer.
The correct answer is $0.10 \text{ g}$

#### Explanation:

A formula for half-life decay is:

$A \left(t\right) = A \left(0\right) {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$

Where ${t}_{\frac{1}{2}}$ is the half-life, t is the elapsed time (in the same time units as the half-life), $A \left(t\right)$ is the amount of the radioactive substance remaining after the elapsed time, $A \left(0\right)$ is the amount of the radioactive substance at the start of the elapsed time.

We are given that $A \left(0\right) = 1.60 \text{ g}$, ${t}_{\frac{1}{2}} = 15 \text{ hrs}$ and $t = 60 \text{ hrs}$

A(60" hrs") = 1.60" g"(1/2)^((60" hrs")/(15" hrs"))

A(60" hrs") = 1.60" g"(1/2)^4

$A \frac{60 \text{ hrs") = (1.60" g}}{16}$

$A \left(60 \text{ hrs") = (0.10" g}\right)$