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How many moles of CaCl2 are required to prepare 2.00 liters of 0.804 molar CaCl2. Your answer should go to two decimal places.



Answer :

To determine how many moles of CaCl2 are required to prepare 2.00 liters of a 0.804 molar (M) CaCl2 solution, follow these steps:

1. Understand the concept of molarity (M):
Molarity is defined as the number of moles of solute (in this case, CaCl2) per liter of solution. It is given by the formula:
[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \][/tex]

2. Rearrange the formula to solve for moles of solute:
[tex]\[ \text{moles of solute} = \text{Molarity (M)} \times \text{volume of solution in liters} \][/tex]

3. Substitute the given values into the formula:
Given:
- Volume of the solution (V) = 2.00 liters
- Molarity (M) = 0.804 M

Hence,
[tex]\[ \text{moles of CaCl2} = 0.804 \, \text{M} \times 2.00 \, \text{L} \][/tex]

4. Perform the multiplication:
[tex]\[ \text{moles of CaCl2} = 1.608 \][/tex]

5. Round the result to two decimal places:
[tex]\[ \text{moles of CaCl2} \approx 1.61 \][/tex]

Therefore, 1.61 moles of CaCl2 are required to prepare 2.00 liters of a 0.804 molar CaCl2 solution.

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