To 220 g of a sixty percent solution of acetic acid, 10 g of the substance and 50 g of water were added.

To 220 g of a sixty percent solution of acetic acid, 10 g of the substance and 50 g of water were added. Determine the new mass fraction.

Given:

m (solution) = 220 g;

w% (CH3COOH) = 60%;

m (CH3COOH) = 10 g;

m (H2O) = 50 g;

Find:

w% (CH3COOH in final solution) -?

Solution:

1) Find the mass of acid in solution:

m (CH3COOH) = 220 g * 0.6 = 132 g;

2) Find the mass of the solution after adding a portion of acid and water:

m (solution) = 220 g + 10 g + 50 g = 280 g;

3) Find the mass of acid in the new solution:

m (CH3COOH) = 132 g + 10 g = 142 g;

4) Find the mass fraction of acetic acid in the resulting solution:

w% (CH3COOH) = m (CH3COOH): m (solution) * 100% = 142 g: 280 g * 100% = 50.7%.

Answer: w% (CH3COOH) = 50.7%.