Question

Question #fc073

Answer 1

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 `"500 mL"` of a `"2.0-M"` sulfuric acid solution from a stock solution that is `"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.

`color(blue)("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 `"1.00 g/mL"`. This means that the density of the stock sulfuric acid solution will be

`"SG" = rho_(H_2SO_4)/rho_"water" implies rho_(H_2SO_4) = "SG" xx rho_"water"`

`rho_(H_2SO_4) = 1.84 * "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

`color(blue)(c = n/V implies n = c * V)`

`n_(H_2SO_4) = "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_"stock" = "60 mL"`

So, to prepare your target solution, you would mix `"60 mL"` of the stock solution with enough water to get the total volume of the resulting solution to `"500 mL"`.

This is equivalent to diluting the `"60-mL"` stock solution sample by a dilution factor equal to `25/3`.