In a #1*mL# volume of solution that is #0.10*mol*L^-1# with respect to #CaCl_2(aq)#, what is the NUMBER of #"ions in solution?"#

1 Answer
Jun 11, 2017

Answer:

There are #3xx10^-4*molxxN_A# ions in solution.......

Explanation:

Now #"concentration"# is given by the quotient:

#"concentration"="Moles of solute"/"Volume of solution"#

And thus to get the #"moles of solute"# we takes the product

#"Volume of solution "xx" concentration"#.......

And so here,

#"moles of calcium chloride"# #=#

#1*mLxx10^-3L*mL^-1xx0.10*mol*L^-1=10^-4*mol#

WITH RESPECT TO THE SALT #CaCl_2#.

Why did we emphasize #"the salt"#? Because EACH formula unit of #CaCl_2# contributes 3 ions to the solution, i.e.....

#CaCl_2(s)stackrel(H_2O)rarrCa^(2+) + 2Cl^-#

And so we have #3xx10^-4*molxx6.022xx10^23*mol^-1=??# ions??