What mass of barium chloride is required to make a 100*cm^3 volume of 0.25*mol*L^-1 solution with respect to BaCl_2(aq)?

1 Answer
Jun 24, 2017

Approx. 5*g.........

Explanation:

"Concentration"="Moles of solute (mol)"/"Volume of solution (L)"

And thus the product "volume"xx"concentration"="moles of solute."

And thus..........

n_(BaCl_2)=0.25*mol*L^-1xx100*cm^3xx10^-3*L*cm^-3

=0.025*mol BaCl_2

And since "no. of moles"="mass"/"molar mass"

"mass"=0.025*cancel(mol)xx208.23*g*cancel(mol^-1)=??g

All I have done here is to use the given quotient that defines "concentration", and divided or multiplied to give the quantity I want. As a check on my arithmetic, I included the units of each quantity. The fact that I clearly got an answer in "grams" for a question that asked for an answer in "grams" persuades me that I got the order of operations right (for once!).