# Question fc073

Dec 6, 2015

Here's what I got.

#### Explanation:

Since I'm not sure what the actual question is, I will assume that you're interested in finding out how to prepare $\text{500 mL}$ of a $\text{2.0-M}$ sulfuric acid solution from a stock solution that is $\text{90% w/w}$ and has a specific gravity equal to $1.84$.

Now, a substance's specific gravity is simply the ratio between that substance's density and the density of a reference substance, usually water.

$\textcolor{b l u e}{\text{SG" = rho_"substance"/rho_"water}}$

Since no mention of temperature was made, I think that you can assume the density of water to be equal to $\text{1.00 g/mL}$. This means that the density of the stock sulfuric acid solution will be

$\text{SG" = rho_(H_2SO_4)/rho_"water" implies rho_(H_2SO_4) = "SG" xx rho_"water}$

${\rho}_{{H}_{2} S {O}_{4}} = 1.84 \cdot \text{1.00 g/mL" = "1.84 g/mL}$

Now, use the molarity and volume of the target solution to determine how many moles of sulfuric acid it must contain

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{{H}_{2} S {O}_{4}} = {\text{2.0 M" * 500 * 10^(-3)"L" = "1.0 moles H"_2"SO}}_{4}$

Use sulfuric acid's molar mass to help you determine how many grams would contain this many moles of sulfuric acid

1.0 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * "98.08 g"/(1color(red)(cancel(color(black)("moles H"_2"SO"_4)))) = "98.08 g"

Use the stock solution's percent concentration by mass to determine what mass of the stock solution would contain this many grams of sulfuric acid

98.08 color(red)(cancel(color(black)("g H"_2"SO"_4))) * "100 g stock solution"/(90color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "109 g stock solution"

Finally, use the stock solution's density to calculate what volume would contain this many grams

109 color(red)(cancel(color(black)("g stock solution"))) * "1 mL"/(1.84color(red)(cancel(color(black)("g stock solution")))) = "59.2mL"#

Now, you need to round this value to one sig fig, the number of sig figs you have for the volume of the target solution.

${V}_{\text{stock" = "60 mL}}$

So, to prepare your target solution, you would mix $\text{60 mL}$ of the stock solution with enough water to get the total volume of the resulting solution to $\text{500 mL}$.

This is equivalent to diluting the $\text{60-mL}$ stock solution sample by a dilution factor equal to $\frac{25}{3}$.