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 solution of 40 % sulfuric acid is required in the digester. The required volume of 40% acid solution is 1500 liters and its specific gravity is 1.3070. . The procedure requires the complete addition of water before the acid. What is the required volume of water and acid to make the 40% solution.

Hello!
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/