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How much CaCl2 is needed to make 200 gal of a 35 percent solution of CaCl2 in a 200 gal tank?
Wiki User
∙ 11y ago
Depends on how you measure your 35% - 35% w/w or 35% w/v.
Also whether you are measuring US gal or Imperial Gal. And what form of calcium chloride you are starting with (fused anhydrous or dihydrate are the two common ones)
Easier to work in litres -
200 gal imperial is about 900 litres.
Density of 35%w/w is around 1.357 @ 25°C so 1 litre of finished 35% w/w CaCl2 solution weighs 1.357 kg.
900 lt of 35%w/w weighs 900 * 1.357 = 1221kg
amount of CaCl2 (anhydrous - 100%) required = 1221 * .35 kg
= 428 kg
Wiki User
∙ 11y ago
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What mas of CaCL2 should be used to make 4.5 L of .320 M CaCL2 solution?
Molarity = moles of solute/Liters of solution 0.320 M CaCl2 = moles CaCl2/4.5 Liters = 1.44 moles of CaCl2 1.44 moles CaCl2 (110.978 grams/ 1 mole CaCl2) = 159.81 grams needed so, considering the sigi figis, 160 grams needed.
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A chemist has one solution that is 60 percent chlorinated and another solution that is 40 percent chlorinated. How much of the first 60 percent is needed to make a 100 L solution that's 50 percent?
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Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!). Example: What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution? Solution: 11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2 100 mL x 1 L / 1000 mL = 0.10 L molarity = 0.10 mol / 0.10 L molarity = 1.0 M http://chemistry.about.com/od/lecturenotesl3/a/concentration.htm http://www.tpub.com/content/MIL-SPEC/MIL-P/MIL-P-71158/MIL-P-7115800013.htm http://www.tpub.com/content/armymedical/md0837/md08370139.htm
