Chapter 14: Problem 71
Calculate the molarity of a solution that contain 15.7 \(\mathrm{g}\) of \(\mathrm{CaCO}_{3}\) dissolved in 275 \(\mathrm{mL}\) of water.
Short Answer
Expert verified
The molarity of the solution containing 15.7 g of CaCO3 dissolved in 275 mL of water is approximately 0.571 M.
Step by step solution
01
Calculate the moles of CaCO3
To find the number of moles of CaCO3, we will use the given mass (15.7 g) and the molar mass of CaCO3. The formula for determining the moles is:
Moles = mass / molar mass
First, determine the molar mass of CaCO3:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (Oxygen has three atoms in CaCO3, so you need to multiply its molar mass by 3)
Molar mass of CaCO3 = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol
Now, we can find the number of moles:
Moles = 15.7 g / 100.09 g/mol
02
Convert volume from mL to L
To find the molarity of the solution, we need to convert the volume from milliliters to liters. Since 1 liter = 1000 milliliters, we can use the following conversion:
Volume in liters = volume in mL / 1000
So, 275 mL = 275 / 1000 = 0.275 L
03
Calculate the molarity of the solution
We can now use the formula for molarity:
Molarity (M) = moles of solute / volume of solution in liters
First, calculate the moles of CaCO3:
Moles = 15.7 g / 100.09 g/mol = 0.1569 moles
Then, find the molarity of the solution:
M = 0.1569 moles / 0.275 L = 0.5705 M
The molarity of the solution is approximately 0.571 M.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Moles of Compound
The concept of moles is fundamental in chemistry as it helps us quantify the amount of a substance. A mole is a basic unit that measures the amount of a chemical substance, based on Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles/mol. To calculate the moles of a compound, you use the formula:
- Moles = \(\frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}}\)
Molar Mass
Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). It provides a bridge between the atomic scale and the macroscopic scale, allowing scientists to count "particles" by weighing them. To calculate the molar mass:
- Find the atomic mass of each element in the compound, usually from the periodic table.
- Multiply the atomic mass of each element by the number of atoms of that element in the compound.
- Add these values together to find the molar mass of the entire compound.
- Calcium (Ca) has a molar mass of 40.08 g/mol.
- Carbon (C) has a molar mass of 12.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol, and since there are three oxygen atoms, we multiply this by 3.
- Adding these gives us the molar mass: 40.08 + 12.01 + (3 \times 16.00) = 100.09 g/mol.
Volume Conversion
Volume conversion is often necessary when working with solutions because concentration calculations typically require volume in liters. Many laboratory settings and data provide volume in milliliters (mL), so it's crucial to be able to convert between these units. The conversion is simple:
- 1 liter = 1000 milliliters
- To convert from mL to L, divide by 1000.
Solution Concentration
The concept of solution concentration is key in chemistry, as it describes how much solute is present in a given amount of solvent. Molarity is a common measure of concentration, expressed in moles per liter (M), and is calculated using the formula:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]In our example, we had previously calculated that there are approximately 0.157 moles of \(\text{CaCO}_3\) in 0.275 L of solution. Therefore, the molarity (concentration) of the solution is:\[ M = \frac{0.157 \text{ moles}}{0.275 \text{ L}} \approx 0.571 \text{ M} \]Understanding molarity is essential for preparing solutions of precise concentrations, necessary in reactions and processes where the concentration impacts the chemical behavior. It provides a way to express concentration that simplifies the math involved in chemical equations and reactions.