Hi, can someone help me with the mixture calculation?

Question

picus48 · Answer

At w=2.5%, 2.5 g of carbonate in 100 g of solution (2.5 g of carbonate+97.5 g of water) are added.

At w=4%, 4 g of carbonate in 100 g of solution (2.5 g of carbonate+97.7 g of water) are obtained.

The invoice is valid for x = 1.6 g:

w = (2.5 + 1.6) g/101.6 g = 0.04035 ==> 4.035%

cg1967 · Answer

Cover:

g of carbonate are contained in 100 g of solution and 4 g are to be obtained. It is necessary to add about 1.5 g to 100 g, i.e. about 3.75 g to 250 g. There are options 4 and 5.

Invoicing:

250 g of solution contain 6.25 g of carbonate on 250 g of solution.

Case 1: +3.4 g of carbonate

9.65 g / 253.4 g = 3.81%

Case 2 +3.9 g carbonate

10.15 g / 253.9 g = 4.00%

tomrichter · Answer

Step 1: We calculate how much soda (here is, in chemistry is rather rare, the mass and not the amount of substance in demand) is already in solution:

m1 = w1 * M1

with the conventions that m should be the mass soda, M the mass solution, index 1 before the addition and index 2 after the addition.

The same applies

m2 = w2 * M2

and if we call the mass of the desired addition x, it becomes:

m1 + x = w2 * (M1 + x)

Except x, all variables are given – simply insert and re-form them by x. You know the transformation from the math class, you need to practice again and again, so if necessary!