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4. In a plant, sulfuric acid is purchased with a concentration of 95%. The specific gravity of the 95% sulfuric acid is 1.8358. A...

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The mass conservation law will help us in this problem. Denote `rho_1` the known density of 95% sulfuric acid, `rho_2` the known density of 40% sulfuric acid (the given numbers are in  `g/(cm^3)` ), `rho_w=1g/(cm^3)`  the (known) density of pure water. Let the unknown volumes of 95% acid and water be `V_a` and `V_w,` and the known volume of resulting solution `V_2.`


Then the mass before the mixing is  `V_w rho_w + V_a rho_1`  and the mass after the mixing is  `V_2 rho_2,` and they must be equal:


`V_w rho_w + V_a rho_1 =V_2 rho_2.`


The mass of pure sulfuric acid also remains the same, thus


`0.95 V_a rho_1 = 0.40 V_2 rho_2.`


From this equation we can find the required volume of 95% acid:  `V_a = 0.40/0.95 *V_2 *rho_2/rho_1.`


Substitute it into the first equation and obtain


`V_w rho_w +0.40/0.95 V_2 rho_2 =V_2 rho_2, or V_w = V_2rho_2/rho_w (1 - 0.40/0.95).`



Numerically the answers are


`V_a = 0.40/0.95 *1500*1.3070/1.8358 approx 450` (liters of 95% acid)


and


`V_w = 1500*1.3070* (1 - 0.40/0.95) approx 1135` (liters of water).


Note that the volume is not preserved.

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