The two concentrations are c) equal.
Let's calculate the concentration by mass of "NaOH" in each solution.
2 mol/L "NaOH"
The density of 2 mol/L "NaOH" is 1.08 g/mL.
∴ The mass of 1 L of the solution is
1000 color(red)(cancel(color(black)("mL solution"))) × "1.08 g solution"/(1 color(red)(cancel(color(black)("mL solution")))) = "1080 g solution"
The mass of "NaOH" in this solution is
2 color(red)(cancel(color(black)("mol NaOH"))) × "40.00 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "80 g NaOH"
% "NaOH (m/m) " = "mass of NaOH "/"mass of solution " × 100 % = (80 color(red)(cancel(color(black)("g"))))/(1080 color(red)(cancel(color(black)("g")))) × 100 % = 7 %
2 mol/kg "NaOH"
Here, you have 80 g of "NaOH" in 1 kg of water.
"Mass of solution" = "80 g + 1000 g" = "1080 g"
% "NaOH (m/m) " = "mass of NaOH "/"mass of solution " × 100 % = (80 color(red)(cancel(color(black)("g"))))/(1080 color(red)(cancel(color(black)("g")))) × 100 % = 7 %
The two concentrations are the same within experimental uncertainty.
