How many milligrams of salt must be added to 18 milligrams of a 20 percent salt solution to obtain a 35 percent salt solution?

Answer 1

this question is not hard to answer, but it does require that one make some assumptions.

The simplest answer assumes that the 20% salt solution refers to per cent by mass. Thus: a 20% salt solution is one which contains 20 mg salt for every 100 mg solution where the solution consists of a mixture of 20 mg salt plus 80 mg water.

A 35% salt solution would contain 35 mg salt for every 65 mg water.

Now, assuming that all the water in 18 mg of a 20% salt solution remains in the final solution we see that 18 mg salt solution x 20 mg salt/100 mg salt solution gives 3.6 mg of salt; thus, there are 18 mg total solution - 3.6 mg salt = 14.4 mg water.

So the final salt solution must be one that contains 14.4 mg water and enough salt to make it 35% salt by mass. Mathematically, this is written as

Z mg salt/(Z mg + 14.4 mg water) = 35/100

This gives Z = 0.35*(Z + 14.4) or

Z = 0.35*Z + 0.35*14.4 which is same as

Z = 0.35*Z + 5.04

and 0.65*Z = 5.04

so Z = 5.04/0.65 = 7.75 mg total salt needed. We started with 3.6 mg salt, so we must add 7.75 -3.6 = 4.15 mg salt

Check: 7.75 mg salt/(7.75 mg salt + 14.4 mg water) = 0.35 or 35%

There you go! --assuming that much salt dissolves that amount of water!