Question

given 1.00L each of 0.200M and 0.400M HCL using only these solutions and assuming that volume are additive what is the maximum volume of 0.275M of HCL that can be prepared.​

Answer 1

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Answer 2

The maximum volume of 0.275 M of HCl prepared from the given solutions is 1.6 liters.

Given,

Two solutions of HCl are given with additive volumes

M₁ = 0.200 M

M₂ = 0.400 M

V₁ = V₂ = 1 L = 1000 mL

To Find,

The maximum volume of 0.275M of HCl that can be prepared.​

Solution,

The required molarity is less than 0.400 M, thus we will add only some amount of this solution and the whole of the 0.200 M solution.

Let the volume of 0.400 M solution mixed be 'x' L.

Thus, the volume of the new solution becomes (1 + x) L

Using the formula -

M₁V₁ + M₂V₂ = M'V'

0.200(1) + 0.400(x) = 0.275 * (1 + x)

0.200 + 0.400x = 0.275 + 0.275x

Multiplying throughout by 1000 we get -

125x = 75

x = $\frac{75}{125}$

x = 0.6

⇒ 1 + x = 1 + 0.6 = 1.6 L

Hence, we get the volume of the solution as 1.6 L.

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