# Thread: Calculate a third mixture from blending two different mixtures

1. ## Calculate a third mixture from blending two different mixtures

Here's the question:

I have an unlimited supply of 1% mixture of salt and water (Mixture #1)

I have 2.5kg of a 10% mixture of salt and water (Mixture #2)

How much of Mixture #1 do I need to add to Mixture #2 in order to make a 3.33% 9000mL solution?

I am lost and can't figure this out!!!!

Assume that the density of both mixtures are equal to the density of water

2. ## Re: Calculate a third mixture from blending two different mixtures

1 kg of water is 1 liter or 1000 ml

so we have $2500 ml$ of 10% mixture

and unlimited supply of 1% mixture

as is this problem either works or it doesn't, let's see.

You need $9000ml$ as the final amount so you must add $9000ml-2500ml=6500ml$ of the 1% mixture to the 10% mixture.

This results in

$6500ml(0.01) + 2500ml(0.1) = 65ml+250ml = 315 ml$ of salt

This results in a $\dfrac{315ml}{9000ml} = 0.035 = 3.5\%$ solution

are you supposed to be able to vary the amount of 10% solution as well?

3. ## Re: Calculate a third mixture from blending two different mixtures

You want 9000 mL that is 3.33% salt so you want 9000(0.033)= 299.7 grams of salt. If you use X mL of the 1% solution and Y =mL of the 10% solution that will contain 0.01X+ 0.10Y grams of salt.

In order that you have 9000 mL of solution you must have X+ Y= 9000. In order that you have 299.7 grams of salt, you must have 0.01X+ 0.10Y= 299.7. Solve those two equations for X and Y.

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