The flask contained 200 g of a solution with a mass fraction of the substance of 16%. Some of the solution was cast

The flask contained 200 g of a solution with a mass fraction of the substance of 16%. Some of the solution was cast off, but the same amount of water was added. In this case, the mass fraction of the substance became equal to 10%. How much water was added?

Given:
m1 solution = 200 g
ω1 (in-islands) = 16% = 0.16
ω3 (in-va) = 10% = 0.1

Find:
m (H2O) -?

Solution:
1) Take the mass of added water as (x) g;

2) Find the mass of the intermediate solution:
Since the mass of the cast solution is equal to the mass of the added water, then
m2 solution = 200 – x g;

3) Find the mass of dry matter in the intermediate solution:
As a result of the reflux of a part of the solution (before adding water), the mass fraction of the substance remained unchanged – 16%. Then
m2 (in-va) = ω2 (in-va) * m2 solution = 0.16 * (200 – x) g;

4) Find the mass of the final solution:
Since the same mass of liquid was cast and added, then
m3 solution = m1 solution = 200 g;

5) Find the mass of dry matter in the final solution:
m3 (in-va) = ω3 (in-va) * m3 solution = 0.1 * 200 = 20 g;

6) Since water was added to the solution, the mass of dry matter in the intermediate and final solutions remained the same. Means
m3 (in-va) = m2 (in-va);
Make and solve the equation:
20 = 0.16 * (200 – x);
x = 75;

m (H2O) = x = 75 g.

Answer: The mass of added water is 75 g.