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How much water must be added to 100 cm3 of 50 g dm-3 hydrochloric acid to make dilute hydrochloric acid with a concentration of 20g gm-3.

Question

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Solution

To solve this problem, we first need to understand the concept of dilution. The formula for dilution is M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

  1. Identify the given values: M1 = 50 g/dm^3 V1 = 100 cm^3 = 0.1 dm^3 (since 1 dm^3 = 1000 cm^3) M2 = 20 g/dm^3 V2 = ?

  2. Substitute the given values into the formula: 50 g/dm^3 * 0.1 dm^3 = 20 g/dm^3 * V2

  3. Solve for V2: V2 = (50 g/dm^3 * 0.1 dm^3) / 20 g/dm^3 V2 = 0.25 dm^3

  4. Convert V2 back to cm^3: V2 = 0.25 dm^3 * 1000 = 250 cm^3

  5. Calculate the amount of water to be added: The amount of water to be added = Final volume - Initial volume = 250 cm^3 - 100 cm^3 = 150 cm^3

So, 150 cm^3 of water must be added to the 100 cm^3 of 50 g/dm^3 hydrochloric acid to make a dilute hydrochloric acid with a concentration of 20 g/dm^3.

This problem has been solved

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