Given a #1*L# volume of solution that is #1.25*mol*L^-1# with respect to #MgCl_2#, there are...? #"a. 2 moles of ions in solution."# #"b. 1.25 moles of ions in solution."# #"c. 2.5 moles of ions in solution."# #"d. 2 moles of ions in solution."#

Given a #1*L# volume of solution that is #1.25*mol*L^-1# with respect to #MgCl_2#, there are...
#"a. 2 moles of ions in solution."#
#"b. 1.25 moles of ions in solution."#
#"c. 2.5 moles of ions in solution."#
#"d. 2 moles of ions in solution."#

1 Answer
anor277
May 3, 2017

#"Option c............"#

Explanation:

The #"mole"#, #N_A#, is simply a number. Admittedly, it is a very large number, #N_A=6.022xx10^23*mol^-1#.

And if you have a mole of stuff, you have #6.022xx10^23# INDIVIDUAL ITEMS of that stuff. Here you have a concentration of #1.25*mol*L^-1# #MgCl_2#. In #1*L# of solution there are #1.25*mol# #Mg^(2+)# ions, BUT ALSO #2xx1.25*mol# #Cl^-# ions.....

And thus in total you have #2.5xxN_A# #Cl^-# per litre of this solution. This is a basic, underlying principle of chemistry, and if you can develop this idea of chemical equivalence, where MASS and CONCENTRATION corresponds to numbers of molecules and atoms, you will save yourself a lot of trouble,

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