Question #30f4f

1 Answer
Jun 18, 2017

Answer:

We WOULD normally use #C_1V_1=C_2V_2#, so.........#C_2~=0.02*mol*L^-1#

Explanation:

The product #C_1V_1# has units #mol*L^-1xxL#, i.e. you get an answer in #mol# as required. Here, we want the #"concentration"#, that is given by the quotient #"moles"/"volume"#..........

And so here we use this quotient.................

#C_1V_1=C_2V_2#, #C_2=(C_1V_1)/V_2#,

#C_2=(50.0*mLxx10^-3*L*mL^-1xx0.195*mol*L^-1)/(5.00xx10^2*mLxx10^-3*L*mL^-1)#

#=1/10C_1#

And this makes sense given that we have diluted the volume #"tenfold"#.