# To a 40 percent solution of hydrochloric acid, 50 g of pure acid was added, after which the concentration

**To a 40 percent solution of hydrochloric acid, 50 g of pure acid was added, after which the concentration of the solution became equal to 60 percent. Find the initial weight of the solution?**

1. Let:

m1 is the mass of a 40% solution;

m2 = 50 g mass of pure hydrochloric acid;

m = 50 g is the mass of the mixed solution;

r1 = 40%, r2 = 100%, r = 60% are the concentrations of the corresponding solutions.

2. We get the equations:

{m1 + m2 = m;

{r1m1 + r2m2 = rm;

{m2 = m – m1;

{r1m1 + r2 (m – m1) = rm;

{m2 = m – m1;

{r1m1 + r2m – r2m1 = rm;

{m2 = m – m1;

{r1m1 – r2m1 = rm – r2m;

{m2 = m – m1;

{(r1 – r2) m1 = (r – r2) m;

{m2 = m – m (r – r2) / (r1 – r2);

{m1 = m (r – r2) / (r1 – r2);

{m2 = m (r – r1) / (r2 – r1);

{m1 = m (r – r2) / (r1 – r2).

3. Let’s divide the equations:

m1 / m2 = (r2 – r) / (r – r1);

m1 = m2 (r2 – r) / (r – r1);

m1 = 50 * (100 – 60) / (60 – 40) = 50 * 40/20 = 100 (g).

Answer: 100 g.