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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.
https://courses.lumenlearning.com/boundless-chemistry/chapter/history-of-atomic-structure/
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