Question

How many milliliters of a $1.75\mathrm{M}\mathrm{LiCl}$ solution contain $15.2\mathrm{g}$ of $\mathrm{LiCl}?$

Short answer

204.9 mL

Step-by-step solution

- Determine the Molar Mass of LiCl

Calculate the molar mass of $\text{LiCl}$. $\text{Li}$ has a molar mass of approximately $\text{6.94 g/mol}$ and $\text{Cl}$ is about $\text{35.45 g/mol}$. Thus, the molar mass of $\text{LiCl}$ is $\text{6.94 g/mol}+\text{35.45 g/mol}=\text{42.39 g/mol}$.

- Convert Grams to Moles

Use the molar mass to convert from grams to moles. Given $\text{15.2 g}$ of $\text{LiCl}$: $$\text{Moles of LiCl}=\frac{\text{15.2 g}}{\text{42.39 g/mol}}\approx 0.3585\text{ moles}$$

- Use the Molarity Formula

Given the molarity $\text{1.75 M}$, the formula to find volume $\text{V}$ is: $\text{M}=\frac{\text{moles of solute}}{\text{volume of solution in liters}}$. Rearrange this to find the volume: $\text{V}=\frac{\text{moles of solute}}{\text{M}}$.

- Calculate the Volume in Liters

Substitute the values into the rearranged formula: $$\text{V}=\frac{0.3585\text{ moles}}{1.75\text{ M}}\approx 0.2049\text{ liters}$$

- Convert Liters to Milliliters

Convert the volume from liters to milliliters by multiplying by 1000: $$0.2049\text{ liters}\times 1000=204.9\text{ mL}$$

Key concepts

Molarity

Molarity is a measure of the concentration of a solution. It tells us how many moles of solute are present per liter of solution. The formula for molarity (M) is: M = $\frac{\text{moles of solute}}{\text{volume of solution in liters}}$.
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

Molar mass is the mass of one mole of a substance (usually in grams). It’s used to convert between grams and moles. You can find the molar mass by summing the atomic masses of all atoms in a molecule.
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

Often, we need to convert units to ensure that our values are correctly matched in calculations. Common conversions include:

• 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

Chemical calculations involve using relationships between quantities like mass, volume, molarity, and moles. The steps to solve such problems are typically:

• 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.