Given #1.0*mol*L^-1# solutions of #NaCl(aq)#, what are the masses of solute in.....?

#(i)# #"a 1 L volume of solution;"#
#(ii)# #"a 250 mL volume of solution;"#
#(iii)# #"a 100 mL volume of solution?"#

1 Answer
Sep 18, 2017

Answer:

Well the molar mass of #NaCl# is #(22.99+35.45)*g*mol^-1=# #58.44*g*mol^-1#.......

Explanation:

From where did I get the atomic masses?

For the remaining problems, we use the relationship.....

#"Concentration/Molarity"="Moles of solute"/"Volume of solution"#

And so, if we gots a #1*L# volume of #1*mol*dm^-3# solution with respect there is a mass of.....

#1*Lxx1*mol*L^-1xx58,44*g*mol^-1=58.44*g#

And note that #1*dm^3-=(1xx10^-1*m)^3-=1xx10^-3*m^3=1*L#, because there are #1000*L# in a #m^3#.

And we also gots #250*cm^3xx10^-3*dm^3*cm^-3*2*mol*dm^-3xx58.44*g*mol^-1=29.22*g#

And I will let you the third.......