The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 4.7 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?

0.94 liters antifreeze is  drained out and replaced with water to get a concentration to 50 per cent of the solution.

Step-by-step explanation:

Total amount of the solution is 4.7 liters.

It contains 70 percent antifreeze = 70% * of 4.7 = 3.29 liters of antifreeze

It contains 30 percent water = 30% * of 4.7= 1.41 liters water

For the solution to have 50 % antifreeze it must contain

4.7 * 50%= 2.35 antifreeze.

We have 3.29 liters and we want to have 2.35 liters of antifreeze

Therefore we will (3.29-2.35 =)0.94 liters drained out and replace it with water to get a concentration to 50 per cent.

water molecules will move from the side of higher water concentration to the side of lower concentration until both solutions are isotonic. at this point, equilibrium is reached. in an isotonic solution, the flow of water in and out of the cell is happening at the same rate.

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step-by-step explanation:

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