A different amount of water was added three times to a solution of salt with a mass of 500 g, as a result

A different amount of water was added three times to a solution of salt with a mass of 500 g, as a result of which the salt concentration was compressed 2 times each time and, as a result, became equal to 4 percent. Find the mass of salt in grams

1. The mass of the salt solution: M = 500 g;

2. The last concentration of salt in the solution is equal to: Kc = 4%;

3. The salt solution was diluted N = 3 times;

4. The primary concentration of salt in the solution was: K = Kc * N ^ 2 = Rc * 8 = 4 * 8 = 32%;

5. The mass of salt in the primary solution was equal to:

Ms = (K / 100) * M = 32/100 * 500 = 160 g;

Мв1 = М – Мс = 500 – 160 = 340 g;

6. For the first dilution of the solution to Kc1 = 16%, X1 g of water was added to it:

MS = (Ks1 / 100) * K1;

K1 = (100 * Ms) / 16 = (100 * 160) / 16 = 1000 g;

X1 = K1 – Mc = 2000 – 160 = 840 g;

7: For the second dilution of the solution to Kc2 = 8%, X2 g of water was added to it:

MS = (Ks2 / 100) * K2;

K2 = (100 * Ms) / 8 = (100 * 160) / 8 = 2000 g;

X2 = K2 – Ms = 2000 – 160 = 1840 g;

8: For the third dilution of the solution to Kc3 = 4%, X3 g of water was added to it:

MS = (Kc3 / 100) * K3;

K3 = (100 * Ms) / 4 = (100 * 160) / 4 = 4000 g;

X3 = K3 – Ms = 4000 – 160 = 3840 g;

9. The concentration of the solution is equal to: Kc4 = (160/4000) * 100 = 4%.

Answer: the mass of salt in the solution was originally equal to 160 grams.