1000g of a 55%(mass percentage) nitric acid solution is to be diluted to 20%(mass percentage) nitric acid solution.how many liters of water should be added to the starting solution?
70%
To solve this problem, we need to determine how many liters of water should be added to the initial 55% nitric acid solution in order to achieve a 20% nitric acid solution. Here are the steps to calculate the answer:
Step 1: Calculate the initial mass of nitric acid in the 1000g solution.
To do this, multiply the mass of the solution (1000g) by the mass percentage of nitric acid (55%):
Mass of nitric acid = 1000g * (55/100) = 550g
Step 2: Calculate the final mass of nitric acid in the diluted solution.
Since we want the final solution to be 20% nitric acid, we can assume that the mass of nitric acid will remain the same. Therefore, the final mass of nitric acid in the diluted solution will also be 550g.
Step 3: Calculate the amount of water needed to achieve the desired concentration.
To do this, subtract the final mass of nitric acid from the initial total mass of the diluted solution:
Amount of water needed = Total mass of diluted solution - Final mass of nitric acid
Since the final mass of nitric acid is 550g and the total mass of the diluted solution is equal to the initial mass of the solution (1000g), we can substitute the values:
Amount of water needed = 1000g - 550g = 450g
Step 4: Convert the mass of water to volume in liters.
To convert grams to liters, divide the mass of water by the density of water. The density of water is approximately 1g/mL or 1g/cm³.
Volume of water needed = Amount of water needed / Density of water
Volume of water needed = 450g / 1g/mL = 450 mL
Since 1 liter is equal to 1000 mL:
Volume of water needed = 450 mL / 1000 mL/L = 0.45 L
Therefore, to dilute the initial 1000g solution of 55% nitric acid to a 20% nitric acid solution, you need to add 0.45 liters of water.