Volume calculation during solution preparation
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The current volume of NaOH is 4.9 m3 with a concentration of 81.23 g/l. It is necessary to prepare 24.5 m3 of alkaline solution with a concentration of 79 g/l. What is the volume of NaOH with a concentration of 410 g/l that needs to be added, and what is the volume of pure water that needs to be added? How to calculate it accurately?
Reply 1# yuning1987: V alkali = (24.5*79 – 4.9*81.23)/410 ≈ 3.75 m³. Assuming a temperature of 20°C, the densities at concentrations of 79 g/l, 81.23 g/l, and 410 g/l are respectively 1.083, 1.085, and 1.339. You can calculate the remaining value by yourself, right? At first glance, it seemed like a middle school-level problem; it’s normal that elementary school students wouldn’t be able to solve it. But later I realized that even high school students would find it quite difficult to solve. . . . Those densities are calculated through joint queries, which is quite troublesome