Question

What mass of solute exists in a `1*L` volume of `0.650*mol*L^-1` `CaBr_2`?

Answer 1

Approx. `130*g`.

Explanation:

`"Concentration"` `=` `"Moles of solute"/"Volume of solution"`, and, clearly, this has units of `mol*L^-1`.

Here we have a `1L` volume of `0.650*mol*L^-1` `CaBr_2(aq)`.

The product,

`"Volume of solution "xx" Concentration"` `=` `"Moles of solute"`

And thus, `1*cancelLxx0.650*mol*cancel(L^-1)=0.650*mol` `CaBr_2`.

And `0.650*cancel(mol)xx199.89*g*cancel(mol^-1)=??*g`