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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`...

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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.

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