Question

What is the decay rate of Mg−28 in the solution after 4.00 days?

A 238 mL sample of an aqueous solution contains 3.05 %MgCl2 by mass. Exactly one-half of the magnesium ions are Mg−28, a beta emitter with a half-life of 21 hours. Given that half the ions are Mg−28, the molar mass of this sample is 96.9 g/mol. Assume a density of 1.02 g/mL for the solution.

Answer 1

I got a decay rate after `"4 days"` to be `2.22 xx 10^(-4) "M/hr"`. Did you get an initial rate of `5.30 xx 10^(-3) "M/hr"`?

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DISCLAIMER: LONG ANSWER!

From the density of the solution, we can find its mass:

`"1.02 g soln"/cancel"mL soln" xx 238 cancel"mL soln" = "242.76 g soln"`,

And the solution contains `3.05%"w/w"` `"MgCl"_2`. Therefore, it contains

`242.76 cancel"g soln" xx "3.05 g MgCl"_2/(100 cancel"g soln") = "7.404 g MgCl"_2`.

Using the molar mass of the sample as `"96.9 g/mol"`, and the assumption that half of the `"Mg"^(2+)` atoms are `"Mg-28"`:

`7.404 cancel("g MgCl"_2) xx cancel("1 mol MgCl"_2)/(96.9 cancel("g MgCl"_2)) xx cancel("1 mol Mg"^(2+))/cancel("1 mol MgCl"_2) xx ("1 mol "_(12)^(28) "Mg"^(2+))/(2 cancel("mol Mg"^(2+))`

`= "0.0382 mols "_(12)^(28) "Mg"^(2+)`

Knowing that its half-life is `"21 hr"`, and that `4` days have passed (or `"96 hr"`), first we find the amount leftover.

Using the integrated rate law for first-order decay in terms of mols:

`ln(n_(""_(12)^(28) "Mg"^(2+))) = -kt + ln(n_(""_(12)^(28) "Mg"^(2+),0))`

The first-order half-life is given by

`t_"1/2" = (ln2)/k`,

thus giving the rate constant as

`=> k = (ln2)/t_"1/2"`.

So:

`ln(n_(""_(12)^(28) "Mg"^(2+))) = -(ln2)(t/t_"1/2") + ln(n_(""_(12)^(28) "Mg"^(2+),0))`

`= -(ln2)("96 hr"/"21 hr") + ln("0.0382 mols "_(12)^(28) "Mg"^(2+))`

`= -6.43`

As a result, this many mols of `""_(12)^(28) "Mg"^(2+)` was left after `4` days:

`n_(""_(12)^(28) "Mg"^(2+)) = e^(-6.43) = "0.00161 mols"`

This was in `"238 mL"`, so its concentration at `t = "96 hr"` is:

`[""_(12)^(28) "Mg"^(2+)] = "0.00161 mols"/"0.238 L" = "0.00672 M"`

Finally, we can write the first-order rate law to find the decay rate at time `t`:

`r(t) = k[""_(12)^(28) "Mg"^(2+)] = (ln2)/(t_"1/2")[""_(12)^(28) "Mg"^(2+)]`

At `t = "96 hr"`, we use the leftover concentration get:

`color(blue)(r(96)) = (ln2)/("21 hr")("0.00672 M")`

`= color(blue)(2.22 xx 10^(-4) "M/hr")`

What was the initial rate of decay? Did you get `5.30 xx 10^(-3) "M/hr"`?