The vessel contains a solution weighing 230 g with a mass fraction of alkali of 12%. Part of the solution was removed

The vessel contains a solution weighing 230 g with a mass fraction of alkali of 12%. Part of the solution was removed from the solution and water was added, the mass of which was 3 times the mass of the removed solution. As a result, the mass fraction of alkali in the solution was 5%. Calculate the mass of the resulting solution.

1. Let the mass of the removed solution:
mdal (solution) = x g;
2.we express the mass of the remaining solution and the mass of the alkali contained in it:
bridge (solution) = mout (solution) – mdal (solution) = 230 – x;
bridge (alkali) = bridge (solution) * w (alkali) = (230 – x) * 0.12;
3.determine the mass of the added portion of water:
madd (H2O) = mdal (solution) * 3 = 3 * x;
4.express the mass of the resulting solution:
mpol (solution) = mr (solution) + madd (H2O) = 230 – x + 3 * x = 230 + 2x;
5.we substitute the obtained expressions into the formula for finding the mass fraction:
wpol (alkali) = mb (alkali): mpol (solution) = ((230 – x) * 0.12): (230 + 2x);
0.05 = (27.6 – 0.12x): (230 + 2x);
11.5 + 0.1x = 27.6 – 0.12x;
0.22x = 16.1;
x = 73.18 g;
6.Calculate the mass of the resulting solution:
mpol (solution) = 230 + 2 * 73.18 = 376.36 g.
Answer: 376.36 g.