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During a water treatment program, 127 grams of calcium nitrate, [tex][tex]$Ca(NO_3)_2$[/tex][/tex], is dissolved in water. The final volume of the solution is 2,300 milliliters. What is the molarity of the solution?

Refer to the periodic table to help you answer. Express your answer to two significant figures.

The molarity of the solution is [tex][tex]$\square$[/tex][/tex] M.



Answer :

GinnyAnswer
To find the molarity of the solution, follow these steps:

1. Convert the volume from milliliters to liters:
- We know that 1,000 milliliters is equivalent to 1 liter.
- So, the volume of the solution in liters is:
[tex]\[ \frac{2300 \text{ milliliters}}{1000} = 2.3 \text{ liters} \][/tex]

2. Determine the molar mass of calcium nitrate [tex]\(\text{Ca(NO}_3\text{)}_2\)[/tex]:
- The atomic masses of the elements are:
- Calcium (Ca): [tex]\(40.08 \text{ g/mol}\)[/tex]
- Nitrogen (N): [tex]\(14.01 \text{ g/mol}\)[/tex] per atom
- Oxygen (O): [tex]\(16.00 \text{ g/mol}\)[/tex] per atom
- Calcium nitrate contains:
- 1 calcium atom
- 2 nitrogen atoms
- 6 oxygen atoms
- Therefore, the molar mass is calculated as:
[tex]\[ 40.08 + (2 \times 14.01) + (6 \times 16.00) = 40.08 + 28.02 + 96.00 = 164.1 \text{ g/mol} \][/tex]

3. Calculate the number of moles of calcium nitrate:
- Using the formula:
[tex]\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
- For the given mass of 127 grams:
[tex]\[ \text{moles} = \frac{127 \text{ grams}}{164.1 \text{ g/mol}} = 0.7739183424741012 \text{ moles} \][/tex]

4. Determine the molarity of the solution:
- Molarity is defined as the number of moles of solute per liter of solution:
[tex]\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \][/tex]
- Substituting the values:
[tex]\[ \text{Molarity} = \frac{0.7739183424741012 \text{ moles}}{2.3 \text{ liters}} = 0.33605841 \text{ M} \][/tex]

5. Round the final answer to two significant figures:
[tex]\[ \text{Molarity} \approx 0.34 \text{ M} \][/tex]

The molarity of the solution is [tex]\(0.34\)[/tex] M.

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