Question #fc073

1 Answer
Dec 6, 2015

Answer:

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#.