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

1 Answer
Jan 5, 2017

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