document.write( "Question 678364: A solution contains 6% salt. How much water should be added to 30 ounces of this solution to make a 1.8% solution? \n" ); document.write( "

Algebra.Com's Answer #421444 by KMST(5328) $\"\"$ $\"About$

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STARTING NOTES:
\n" ); document.write( "I would expect that by \"30 ounces\" of solution the problem means 30 fluid ounces, a measure of volume. It would be weird if it meant that the solution weighed 30 avoidupois ounces (1 pound and 14 ounces).
\n" ); document.write( "It makes me believe we are measuring the volume of water to be added in fuid ounces.
\n" ); document.write( "It also suggests that the solution is a 6% w/v (weight in volume) solution, containing 6 grams of salt per 100 milliliters of solution.
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\n" ); document.write( "THE PROBLEM
\n" ); document.write( "What follows may not be how your teacher expects you to solve the problem, because it is done differently in different classes and books.
\n" ); document.write( "On diluting, the concentration is reduced by the dilution factor, so if you double the initial volume, you halve the concentration; if you triple the initial volume, the final concentration is one third of the initial concentration, and so on.
\n" ); document.write( "The dilution factor is the ratio of volumes, (final volume)/(initial volume).
\n" ); document.write( "The dilution factor is the same as the ratio of concentrations, (initial concentration)/(final cocentration).
\n" ); document.write( "To get the concentration from 6% to 1.8%, the dilution factor needed is
\n" ); document.write( " $\"6%2F1.8=60%2F18=10%2F3=100%2F30\"$ .
\n" ); document.write( "(Those ratios are the same. It's the same proportion)
\n" ); document.write( "So you need to go from 30 ounces to a final volume of 100 ounces.
\n" ); document.write( "In math problems, as you dilute a solution volumes are additive, so if $\"x\"$ ounces of water are added to 30 ounces of solution, the total volume of the diluted solution would be $\"30%2Bx=100\"$ ounces.
\n" ); document.write( " $\"30%2Bx=100\"$ --> $\"30%2Bx-30=100-30\"$ --> $\"highlight%28x=70%29\"$
\n" ); document.write( "so $\"highlight%2870%29\"$ ounces of water should be added.
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\n" ); document.write( "THE SCIENCE RANT:
\n" ); document.write( "The name ounce is applied to differenty units of force, mass (avoirdupois ounce, troy/apothecary ounce) and volume (fluid ounce), and makes for some confusion.
\n" ); document.write( "Scientists do not use ounces, and specify when in doubt. (If a scientist says a metric ton, I know it is 1000 kilograms, but an Iowa farmer who says a ton probably means 2000 pounds, which is not the same).
\n" ); document.write( "As an analytical chemist, I would probably express the concentration of a salt solution as 6% w/v (weight in volume) meaning 6 gram salt in 100 milliliters of solution.
\n" ); document.write( "On rare occasions, I would be giving a concentration in terms of weight in weight, as in 6% w/w, meaning 6 grams of salt in 100 g of solution, which would have a volume of about 96 milliliters, because the density would be 1.04 gram per milliliter. So 6.0% w/w =6.2% w/v.
\n" ); document.write( "Volumes are not exactly additive in the real world, although in may cases they are almost exactly additive.
\n" ); document.write( "The density of a salt solution changes a bit with concentration, and also with temperature. At $\"20%5Eo\"$ Celsius, density is 1.04 g/mL for salt solutions at concentrations of about 6.0% and 6.2% w/v. It is 1.01 g/mL for 1.8% w/v salt solutions.
\n" ); document.write( "In the case of this problem, adding 70.0 fluid ounces of water to 30.0 fluid ounces of 6.000% w/v solution would yield 100.2 fluid ounces of a solution that would not be exactly 1.800% w/v, but the difference is too small to matter to a nurse diluting a solution. \n" ); document.write( "

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