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.