Saline Dilution With Whole-Gram Constraints
A student research assistant in a materials lab has two stock bottles of saline solution. Bottle A is 18% salt by mass and Bottle B is 6% salt by mass. The protocol calls for 500 grams of a 10% solution for the first batch and 300 grams of a 14% solution for a second batch. The lab has 400 grams of Bottle A available and essentially unlimited Bottle B. A digital balance reports mass to the nearest gram, so the assistant must measure whole-gram amounts; rounding up or down changes the final concentration slightly. Leftover mixed solution cannot be returned to either stock bottle, and the second batch must be made after the first. The assistant wants to know whether both batches are feasible and how much of the 14% batch could be made if the first batch uses whole-gram amounts.
1. Determine whole-gram masses of A and B that make the 500 g batch as close as possible to 10% salt. Show the equation and verify the concentration.
Read the worked answer
Let x be grams of A and y be grams of B. The mass equation is x + y = 500, and the salt equation is 0.18x + 0.06y = 0.10(500) = 50.
Substitute y = 500 - x: 0.18x + 0.06(500 - x) = 50 0.12x + 30 = 50 0.12x = 20 x = 166.67 g
Whole grams give two nearby choices. Using 167 g A and 333 g B: salt = 0.18(167) + 0.06(333) = 30.06 + 19.98 = 50.04 g, so concentration = 50.04/500 = 10.008%. Using 166 g A and 334 g B: salt = 29.88 + 20.04 = 49.92 g, so concentration = 9.984%. The 167/333 split is closer to 50 g of salt, so it is the better whole-gram choice.
2. Using 167 g of A from the first batch, can the assistant make 300 g of 14% solution? If so, how much A and B are needed, and how much A remains?
Read the worked answer
A 300 g batch at 14% needs 0.14(300) = 42 g of salt. Let a be grams of A; then B is 300 - a.
0.18a + 0.06(300 - a) = 42 0.12a + 18 = 42 0.12a = 24 a = 200 g
So the second batch needs 200 g A and 100 g B. Total A used = 167 + 200 = 367 g. Since 367 g is less than the 400 g available, the batch is feasible, and 400 - 367 = 33 g of A remains. The whole-gram amounts are exact here, so no rounding correction is needed.
3. If the first batch is 167 g A and 333 g B, what is the largest whole-gram mass of 14% solution the assistant can make with the remaining A, assuming unlimited B? Round to a feasible whole-gram total and state any concentration tradeoff.
Read the worked answer
After the first batch, 400 - 167 = 233 g of A remains. To maximize the total mass at 14%, use all 233 g of A because A is the more concentrated stock. Let m be the total mass; then B is m - 233, and the salt equation is:
0.18(233) + 0.06(m - 233) = 0.14m 41.94 + 0.06m - 13.98 = 0.14m 27.96 = 0.08m m = 349.5 g
A 349 g batch uses 233 g A and 116 g B: salt = 41.94 + 6.96 = 48.90 g, giving 48.90/349 = 14.011%. A 350 g batch uses 233 g A and 117 g B: salt = 41.94 + 7.02 = 48.96 g, giving 48.96/350 = 13.989%. If the protocol requires at least 14%, the largest feasible whole-gram total is 349 g; 350 g is very close but slightly under 14%.