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What is the concentration of a solution made by diluting 35 mL of 6.0 M HCl to a final volume of 750 mL?


If 2.49g of CuNO3 is dissolved in water to make O.830M solution, what is the volume of the solution in milliliters?


How many moles of NaOH are present in 13.5mL of 0.17 M NaOH?

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Answer:

1. 0.28M

2. 24mL

3. 0.002295moles of NaOH.

Explanation:

1. The 6.0M HCl solution is diluted from 35mL to 750mL, that is:

750mL / 35mL = 21.43 times

The 6.0M HCl solution is diluted 21.43 times, the final concentration is:

6.0M / 21.43 = 0.28M

2. First, we need to convert grams of CuNO3 to moles and then, as molarity is moles per liter we can solve for volume of solution in liters and convert it to mililiters:

Moles CuNO3 -Molar mass 124.9g/mol:

2.49g * (1mol / 124.9g) = 0.020 moles of CuNO3

Volume is:

0.020 moles * (1L / 0.830mol) = 0.0240L * (1000mL / 1L) = 24mL

3. In 13.5mL = 0.0135L of 0.17M NaOH the moles are:

0.0135L * (0.17mol / L) = 0.002295moles of NaOH

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