It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% solution. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?

17.5 litres removed and 17.5 litres of pure antifreeze added

Explanation:

Let k equal the amount of the solution to be removed

amount of antifreeze to be added

0.2(70 - x) + x = 0.4(70)

14 - 0.2x +x = 28

0.8x = 28 -14

x = 14/0.8 = 17.5

x = 17.5 litres removed and 17.5 litres of pure antifreeze added

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

with this one i know the answer in craters one reason i looked it up and i learned about it there is my evidence