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

Answer:

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!).