Chapter 7: Problem 111
How many milliliters of a \(1.75 \mathrm{M} \mathrm{LiCl}\) solution contain \(15.2 \mathrm{~g}\) of \(\mathrm{LiCl} ?\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity
To compute molarity:
- Identify the moles of solute.
- Measure the volume of the solution in liters.
For example, if you know there are 0.5 moles of NaCl in 1 liter of solution, the molarity is 0.5 M. Molarity is crucial for preparing solutions of precise concentration, which is often needed in chemical reactions and laboratory experiments. Molarity helps standardize how we describe concentrations no matter the amount of solution we have.
Molar Mass
For Lithium Chloride (LiCl), we do the following:
- Find the atomic mass of Li (approximately 6.94 g/mol)
- Find the atomic mass of Cl (approximately 35.45 g/mol)
- Add these together: 6.94 g/mol + 35.45 g/mol = 42.39 g/mol
This means that one mole of LiCl weighs 42.39 grams. Molar mass is important in converting between mass and moles, which is necessary for stoichiometric calculations in chemical reactions.
Conversion of Units
- Grams to moles: Use the molar mass (e.g., \( \frac{15.2 \text{ g LiCl}}{42.39 \text{ g/mol}} \approx 0.3585 \text{ moles} \)).
- Liters to milliliters: Multiply by 1000 (e.g., 0.2049 liters \( \times 1000 = 204.9 \text{ mL} \)).
- Moles to number of particles: Use Avogadro's number (6.022\( \times \)10²³ particles/mol).
Conversions ensure that consistent units are used throughout calculations, preventing errors and allowing different measurements to be combined.
Chemical Calculations
- Find the molar mass of the compounds involved.
- Convert the given quantities to the needed units, such as grams to moles.
- Apply the molarity formula: \( M = \frac{moles}{liters} \).
- Rearrange formulas when necessary to find unknown values.
For instance, to find the volume of a LiCl solution containing 15.2 grams of LiCl at 1.75 M, follow these steps:
- Calculate moles: \[ \frac{15.2 \text{ g}}{42.39 \text{ g/mol}} \approx 0.3585 \text{ moles} \]
- Use the molarity formula: \( V = \frac{\text{moles of solute}}{M} \)
- Substitute: \[ V = \frac{0.3585 \text{ moles}}{1.75 \text{ M}} \approx 0.2049 \text{ liters} \]
- Convert to mL: 0.2049 liters \( \times 1000 = 204.9 \text{ mL} \)
Accurate chemical calculations are essential in chemistry to predict the outcomes of reactions and prepare solutions.