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

1 Answer
Jan 5, 2017

Answer:

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#