# Question 74d8d

Aug 12, 2016

The two concentrations are c) equal.

Let's calculate the concentration by mass of $\text{NaOH}$ in each solution.

2 mol/L $\text{NaOH}$

The density of 2 mol/L $\text{NaOH}$ 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 $\text{NaOH}$ 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 $\text{NaOH}$

Here, you have 80 g of $\text{NaOH}$ in 1 kg of water.

$\text{Mass of solution" = "80 g + 1000 g" = "1080 g}$

% "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.