# Question 3c767

Jun 27, 2017

See below.

#### Explanation:

The correct density of concentrated sulfuric acid is ${\text{1.84 g/cm}}^{3}$.

Step 1. Calculate the molar concentration of the conc. ${\text{H"_2"SO}}_{4}$

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

$\text{Mass of conc. acid" = 1000 color(red)(cancel(color(black)("cm"^3))) × "1.84 g"/(1 color(red)(cancel(color(black)("cm"^3)))) = "1840 g}$

${\text{Mass of H"_2"SO"_4 = 1840 color(red)(cancel(color(black)("g conc."))) × (98 "g H"_2"SO"_4)/(100 color(red)(cancel(color(black)("g conc.")))) = "1803 g H"_2"SO}}_{4}$

${\text{Moles of H"_2"SO"_4 = 1803 color(red)(cancel(color(black)("g H"_2"SO"_4))) × ("1 mol H"_2"SO"_4)/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "18.38 mol H"_2"SO}}_{4}$

$c = {\text{18.38 mol"/"1 dm"^3 = "18.38 mol/dm}}^{3}$

Step 2. Calculate the volume of conc. ${\text{H"_2"SO}}_{4}$ 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,

${c}_{1} = {\text{0.054 mol/dm"^3; V_1 = "0.500 dm}}^{3}$
c_2 = "18.38 mol/dm"^3; V_2 = ?#

${V}_{2} = {\text{0.500 dm"^3 × (0.054 color(red)(cancel(color(black)("mol/dm"^3))))/(18.38 color(red)(cancel(color(black)("mol/dm"^3)))) = "0.0015 dm"^3 = "1.5 cm}}^{3}$

You would place about ${\text{200 cm}}^{3}$ of distilled water in a ${\text{500 cm}}^{3}$ volumetric flask, slowly add 1.5 mL of the conc. ${\text{H"_2"SO}}_{4}$ from a pipet with swirling, and then add enough distilled water to make the solution up to the ${\text{500 cm}}^{3}$ mark.