# (a) What is the molar concentration of a concentrated solution that contains 36.00 % "HCl" by mass and has a density of "1.24 g/dm"^3". (b) How you would prepare "1.00 dm"3 of 0.125 mol/L "HCl" from the concentrated solution?

Jun 28, 2017

See below.

#### Explanation:

The correct density of concentrated hydrochloric acid is ${\text{1.19 g/cm}}^{3}$.

Step 1. Calculate the molar concentration of the conc. $\text{HCl}$

Assume that you have ${\text{1 dm}}^{3}$ of the concentrated acid.

$\text{Mass of conc. HCl" = 1000 color(red)(cancel(color(black)("cm"^3))) × "1.19 g"/(1 color(red)(cancel(color(black)("cm"^3)))) = "1190 g}$

$\text{Mass of HCl" = 1190 color(red)(cancel(color(black)("g conc."))) × (36.00 "g HCl")/(100 color(red)(cancel(color(black)("g conc.")))) = "428.4 g HCl}$

$\text{Moles of HCl" = 428.4 color(red)(cancel(color(black)("g HCl"))) × ("1 mol HCl")/(35.5 color(red)(cancel(color(black)("g HCl")))) = "12.07 mol HCl}$

$c = {\text{12.07 mol"/"1 dm"^3 = "12.07 mol/dm}}^{3}$

Step 2. Calculate the volume of conc. ${\text{HCl}}_{2}$ required

We can use the dilution formula for this calculation.

color(blue)(bar(ul(|color(white)(a/a)c_1V_1 = c_2V_2color(white)(a/a)|)))"

where

${c}_{1}$ and ${c}_{2}$ are the molar concentrations and
${V}_{1}$ and ${V}_{2}$ are the corresponding volumes.

We can rearrange this formula to get

V_2 = V_1 × c_1/c_2

In this problem,

${c}_{1} = {\text{0.215 mol/dm"^3; V_1 = "1 dm}}^{3}$
c_2 = "12.07 mol/dm"^3; V_2 = ?

${V}_{2} = {\text{1 dm"^3 × (0.215 color(red)(cancel(color(black)("mol/dm"^3))))/(12.07 color(red)(cancel(color(black)("mol/dm"^3)))) = "0.001 78 dm"^3 = "17.8 cm}}^{3}$

You would place about ${\text{400 cm}}^{3}$ of distilled water in a ${\text{1 dm}}^{3}$ volumetric flask, slowly add 17.8 mL of the conc. $\text{HCl}$ from a pipet with swirling, and then add enough distilled water to make the solution up to the ${\text{1 dm}}^{3}$ mark.