Determine the end of the solution of soda weight 11 liter of solution has 100 grams of salt.

Determine the end of the solution of soda weight 11 liter of solution has 100 grams of salt. How much salt and water should be taken to prepare 20 liters of a 20% solution.

(a solution of soda weighing 11 liters – if the condition is a solution of water with a volume of 11 liters, then the solution is this)

ω = m1 / m * 100%.

m1 is the mass of the solute,

m is the total mass of the solution.

m1 = 100 g;

m = m (salt) + m (water) = 100 + 11000 = 11100 g;

ω = 100 / (11000 + 100) * 100% = 100/11100 * 100% = 0.9%.

To prepare 20 liters of a 20% solution:

In 20 liters of a solution of 20% salt, 80% water;

20 l * 0.8 = 16 l;

16 l = 16000 g;

m H20 = 16000 g in 80% solution;

m salt = 16000 / 0.8 * 0.2 = 400 g.