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You will need more data about all densities (in kg/Litre)

and you must be sure of using mass% = (g solute)/(100 g solution)

Solve two equations for both X and Y:

  1. 4*d50*50 = X*d20*20 + Y*d70*70 (based on salt mass balance in diff. sol'n.)
  2. 4*d50 = X*d20 + Y*d70 (based on solutions mass balance)

In which:

dm = density of the 'm'% salt solution in kg/Litre)

X and Y = volume of the 20% and 70% salt solutions respectively

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Related Questions

A chemist has one solution that is 60 percent chlorinated and another solution that is 40 percent chlorinated How much of the first 60 percent solution is needed to make a 100 L solution that is 50?

To create a 50% chlorinated solution from the 60% and 40% solutions, the chemist will need to mix the two in equal amounts. Therefore, 50 L of the 60% solution and 50 L of the 40% solution are needed to make a 100 L solution that is 50% chlorinated.


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some liquid volumes are not additive, leading to potentially confusing final solution volumes.


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