Question

In a `1*mL` volume of solution that is `0.10*mol*L^-1` with respect to `CaCl_2(aq)`, what is the NUMBER of `"ions in solution?"`

Answer 1

There are `3xx10^-4*molxxN_A` ions in solution.......

Explanation:

Now `"concentration"` is given by the quotient:

`"concentration"="Moles of solute"/"Volume of solution"`

And thus to get the `"moles of solute"` we takes the product

`"Volume of solution "xx" concentration"`.......

And so here,

`"moles of calcium chloride"` `=`

`1*mLxx10^-3L*mL^-1xx0.10*mol*L^-1=10^-4*mol`

WITH RESPECT TO THE SALT `CaCl_2`.

Why did we emphasize `"the salt"`? Because EACH formula unit of `CaCl_2` contributes 3 ions to the solution, i.e.....

`CaCl_2(s)stackrel(H_2O)rarrCa^(2+) + 2Cl^-`

And so we have `3xx10^-4*molxx6.022xx10^23*mol^-1=??` ions??