Question

Question #6c48c

Answer 1

You have `3.558 * 10^(22)` atoms of Ne-22 in that much neon.

The first thing you need to do is determine the abundance of each of the two isotopes. Since you're only dealing with two isotopes, you can express the abundance of one as x and the abundance of the other as 1 - x.

This means that you'll get

`19.99 * x + 21.99 * (1-x) = 21.18`

where

`x` - the decimal abundance of Ne-20;
`(1-x)` - the decimal abundance of Ne-22.

Solve for `x` to get

`19.99 x + 21.99 - 21.99 x = 21.18`

`2x = 1.81 => x = 0.905`

Ne-20 will have a percent abundance of 90.5% and Ne-22 will have a percent abundance of 9.5%.

SIDE NOTE You get percent abundance by multiplying the decimal abundance by 100.

So, for 1 mole of neon, you get 0.905 moles of Ne-20 and 0.095 moles of Ne-22.

Determine how many moles of neon you have in 12.55 g by using its molar mass

`12.55cancel("g Ne") * "1 mole Ne"/(21.18cancel("g Ne")) = "0.6219 moles Ne"`

This means that you get

`0.6216cancel("moles Ne") * "0.095 moles Ne-22"/(1cancel("mole Ne")) = "0.05908 moles Ne-22"`

You know that 1 mole of any substance contains exactly `6.022 * 10^(23)` molecules or atoms of that substance - this is know as Avogadro's number - so the number of atoms present in that many moles of Ne-22 will be

`0.05908cancel("moles Ne-22") * (6.022 * 10^(23)"atoms of Ne-22")/(1cancel("mole Ne-22")) = 3.5578 * 10^(22)"atoms of Ne-22"`

Rounded to four sig figs, the answer will be

`color(green)(3.558 * 10^(22)"atoms of Ne-22")`