Problem 14
Question
What volume of a \(0.580-M\) solution of \(\mathrm{CaCl}_{2}\) contains \(1.28 \mathrm{g}\) solute?
Step-by-Step Solution
Verified Answer
The volume of a \( 0.580\,\mathrm{M} \) solution of CaCl₂ that contains 1.28 g solute is approximately 19.88 mL.
1Step 1: Calculate the molar mass of CaCl₂
To find the moles of CaCl₂, we first need to determine its molar mass. CaCl₂ consists of one calcium (Ca) atom, and two chlorine (Cl) atoms. We will look up their atomic masses on the periodic table and sum them up to find the molar mass of CaCl₂.
Molar mass of Ca = 40.08 g/mol
Molar mass of Cl = 35.45 g/mol
Molar mass of CaCl₂ = (40.08) + 2 × (35.45) = 40.08 + 70.90 = 110.98 g/mol
2Step 2: Calculate the moles of CaCl₂ in the 1.28 g solute
Now, we'll use the molar mass of CaCl₂ to convert the mass of solute (1.28 g) into moles.
moles of CaCl₂ = (mass of solute)/(molar mass of CaCl₂) = \( \frac{1.28\,\mathrm{g}}{110.98\,\mathrm{g/mol}} \) = 0.01153 mol
3Step 3: Use the molarity formula to find the volume of the solution
The molarity formula relates moles, volume, and concentration:
Molarity (M) = \( \frac{moles\,of\,solute}{volume\,of\,solution\,(L)} \)
We are given the molarity (0.580 M) and the moles of solute (0.01153 mol). We will solve for the volume of the solution (in liters).
0.580 M = \( \frac{0.01153\,\mathrm{mol}}{volume\,(L)} \)
Now, we'll solve for the volume:
volume (L) = \( \frac{0.01153\,\mathrm{mol}}{0.580\,\mathrm{M}} \) = 0.01988 L
4Step 4: Convert the volume from liters to milliliters
To make the volume more practical, we'll convert it from liters to milliliters (mL) using the conversion factor:
1 L = 1,000 mL
volume (mL) = 0.01988 L × 1,000 mL/L = 19.88 mL
The volume of a 0.580 M solution of CaCl₂ that contains 1.28 g solute is approximately 19.88 mL.
Key Concepts
Moles CalculationVolume ConversionMolecular Weight Determination
Moles Calculation
To understand moles calculation, let's begin by defining what a "mole" is in chemistry. A mole is a unit of measurement used to express amounts of a chemical substance, analogous to a dozen, but instead of 12, it's Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities. In our example exercise, we're tasked with finding how many moles are in 1.28 grams of \(\mathrm{CaCl}_{2}\). The formula we use is the mass of the substance divided by its molar mass (moles = mass/molar mass).For \(\mathrm{CaCl}_{2}\), we've calculated its molar mass to be 110.98 g/mol. Knowing this, we can determine how many moles 1.28 grams of \(\mathrm{CaCl}_{2}\) would be:
- Calculate moles of \(\mathrm{CaCl}_{2}\): \(\mathrm{moles} = \frac{1.28\,\mathrm{g}}{110.98\,\mathrm{g/mol}} = 0.01153\,\mathrm{mol}\).
Volume Conversion
Volume conversion is critical when dealing with solutions, as it allows us to express volumes in the most convenient units for calculations. In chemistry, solutions are often measured in liters, but practical laboratory or real-world applications often require conversion to milliliters.Continuing with our exercise, after finding the volume in liters from the molarity formula, we convert it to milliliters for a more "real-world" context. The molarity formula is \(M = \frac{moles}{volume\,(L)}\). From this, we solved:
- Volume \((\mathrm{L}) = \frac{0.01153\,\mathrm{mol}}{0.580\,\mathrm{M}} = 0.01988\,\mathrm{L}\).
- Conversion formula: \(1\,\mathrm{L} = 1,000\,\mathrm{mL}\). Hence, \(0.01988\,\mathrm{L} \times 1,000\,\mathrm{mL/L} = 19.88\,\mathrm{mL}\).
Molecular Weight Determination
Molecular weight, also known as molar mass, is indispensable in chemistry, as it's used to convert grams to moles—a necessary step in many chemical calculations.To determine the molecular weight of a compound like \(\mathrm{CaCl}_{2}\), you follow these steps:
- Identify each element's atomic mass from the periodic table.
- For \(\mathrm{CaCl}_{2}\), which consists of one calcium \((\mathrm{Ca})\) and two chlorine \((\mathrm{Cl})\) atoms:
- Calcium: 40.08 g/mol
- Chlorine: 35.45 g/mol (each, hence \(2 \times 35.45\))
- Sum up all the atomic masses to get the molar mass of the compound: \(40.08 + 2 \times 35.45 = 110.98\,\mathrm{g/mol}\).
Other exercises in this chapter
Problem 12
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What volume of \(0.25 \mathrm{M}\) HCl solution must be diluted to prepare 1.00 L of \(0.040 M\) HCl?
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Calculate the sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bic
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Write equations showing the ions present after the following strong electrolytes are dissolved in water. a. \(\mathrm{HNO}_{3}\) b. \(\mathrm{Na}_{2} \mathrm{SO
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