What NaCl concentration results when 204 mLof a 0.760 M NaCI solution is mixed with 417 mL of a 0.430 M NaCl solution?

1 Answer
Aug 30, 2016

Answer:

#[NaCl]# #~=# #0.5*mol*L^-1#

Explanation:

#"Solution 1"#, #"moles of NaCl"# #=# #204xx10^-3*Lxx0.760*mol*L^-1=0.155*mol#.

#"Solution 2"#, #"moles of NaCl"# #=# #417xx10^-3*Lxx0.430*mol*L^-1=0.179*mol#.

Since we must consider the volumes to be additive when these solutions are mixed, we have the following concentration with respect to #NaCl#.

#"Concentration"# #=# #((0.155+0.179)*mol)/((204+417)xx10^-3*L)#

#~=# #0.5*mol*L^-1# with respect to #NaCl#.