Question

Given a `1*L` volume of solution that is `1.25*mol*L^-1` with respect to `MgCl_2`, there are...? `"a. 2 moles of ions in solution."` `"b. 1.25 moles of ions in solution."` `"c. 2.5 moles of ions in solution."` `"d. 2 moles of ions in solution."`

Given a `1*L` volume of solution that is `1.25*mol*L^-1` with respect to `MgCl_2`, there are...
`"a. 2 moles of ions in solution."`
`"b. 1.25 moles of ions in solution."`
`"c. 2.5 moles of ions in solution."`
`"d. 2 moles of ions in solution."`

Answer 1

`"Option c............"`

Explanation:

The `"mole"`, `N_A`, is simply a number. Admittedly, it is a very large number, `N_A=6.022xx10^23*mol^-1`.

And if you have a mole of stuff, you have `6.022xx10^23` INDIVIDUAL ITEMS of that stuff. Here you have a concentration of `1.25*mol*L^-1` `MgCl_2`. In `1*L` of solution there are `1.25*mol` `Mg^(2+)` ions, BUT ALSO `2xx1.25*mol` `Cl^-` ions.....

And thus in total you have `2.5xxN_A` `Cl^-` per litre of this solution. This is a basic, underlying principle of chemistry, and if you can develop this idea of chemical equivalence, where MASS and CONCENTRATION corresponds to numbers of molecules and atoms, you will save yourself a lot of trouble,