Cooking Measurements

# How many mg in a pound of salt?

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## Related Questions

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

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!