Another 28 g of pure alkali was added to a solution of alkali in water. The result is 16% – a percentage

Another 28 g of pure alkali was added to a solution of alkali in water. The result is 16% – a percentage solution weighing 800g. How many grams of alkali was in the original solution?

1. Let’s designate:
m =? – the initial mass of the alkali solution;
p is the percentage of alkali in the solution;
Δm = 28 g – added pure alkali;
m ‘= 800 g is the mass of the solution after addition;
p ‘= 16% is the percentage of alkali after addition.
2. Let’s compose the equations for the amounts of solution and pure alkali in it:
{m ‘= m + Δm;
{m’p ‘= mp + Δm;
{m = m ‘- Δm;
{mp = m’p ‘- Δm;
{m = 800 – 28;
{mp = 800 * 16/100 – 28;
{m = 772 (g solution);
{mp = 100 (g of alkali).
Answer. The initial solution contained 100 g of pure alkali.