Question #fc073

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Question #fc073

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Answer 1 · 2015-12-06T13:27:45

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 $500 mL$ $"500 mL"$ of a $2.0-M$ $"2.0-M"$ sulfuric acid solution from a stock solution that is $90% w/w$ $"90% w/w"$ and has a specific gravity equal to $1.84$ $1.84$ .

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

$SG = \frac{ρ_{substance}}{ρ_{water}}$ $color(blue)("SG" = rho_"substance"/rho_"water")$

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

$SG = \frac{ρ_{H_{2} S O_{4}}}{ρ_{water}} \Rightarrow ρ_{H_{2} S O_{4}} = SG \times ρ_{water}$ $"SG" = rho_(H_2SO_4)/rho_"water" implies rho_(H_2SO_4) = "SG" xx rho_"water"$

$ρ_{H_{2} S O_{4}} = 1.84 \cdot 1.00 g/mL = 1.84 g/mL$ $rho_(H_2SO_4) = 1.84 * "1.00 g/mL" = "1.84 g/mL"$

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

$c = \frac{n}{V} \Rightarrow n = c \cdot V$ $color(blue)(c = n/V implies n = c * V)$

$n_{H_{2} S O_{4}} = 2.0 M \cdot 500 \cdot 10^{- 3} L = {1.0 moles H}_{2} {SO}_{4}$ $n_(H_2SO_4) = "2.0 M" * 500 * 10^(-3)"L" = "1.0 moles H"_2"SO"_4$

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 $25/3$ .

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Question #fc073

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