Question

What is the mass of solute in \(24.5 \mathrm{~mL}\) of \(0.755 \mathrm{M} \mathrm{CoCl}_{2}\) ?

Short answer

The mass of solute in the solution is \(2.401\, \mathrm{g}\).

Step-by-step solution

Identify the Known Values

We are given that the molarity of the solution is \(0.755\, \mathrm{M}\) and the volume of the solution is \(24.5\, \mathrm{mL}\). We need to convert the volume to liters because molarity is expressed in terms of liters. So, \(24.5\, \mathrm{mL} = 0.0245\, \mathrm{L}\).

Calculate Moles of Solute

Using the molarity formula \(M = \frac{n}{V}\) where \(n\) is the number of moles, and \(V\) is the volume in liters, we can find \(n = M \times V = 0.755\, \mathrm{mol/L} \times 0.0245\, \mathrm{L} = 0.0184975\, \mathrm{mol}\).

Convert Moles to Mass

To find the mass of the solute, we need the molar mass of \(\mathrm{CoCl}_{2}\). The molar mass of \(\mathrm{CoCl}_{2}\) is \(58.93\, \mathrm{g/mol}\) for Co plus \(2 \times 35.45\, \mathrm{g/mol}\) for Cl, which totals \(129.83\, \mathrm{g/mol}\). The mass of \(\mathrm{CoCl}_{2}\) is then \(0.0184975\, \mathrm{mol} \times 129.83\, \mathrm{g/mol} = 2.401\, \mathrm{g}\).

Key concepts

Molarity

Molarity is a key concept in chemistry, particularly when you're dealing with solutions. It's a way to express the concentration of a solute in a given solution. Simply put, molarity tells you how many moles of solute are present in one liter of solution. This is vital in stoichiometry because it allows chemists to accurately define the properties of a solution.

To understand molarity, remember that one mole is a specific number of particles, namely Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles. Molarity is then expressed in moles per liter (\( ext{mol/L}\) or simply \( ext{M}\)). When you know the molarity of a solution, you can easily calculate how much solute is present in any given volume. For example, if you have a 0.755 M solution of \( ext{CoCl}_2\), there are \(0.755\) moles of \( ext{CoCl}_2\) in every liter of that solution.

Remember that volume must always be in liters for molarity calculations. So, you may need to convert milliliters to liters by dividing by 1,000, as shown in the original exercise.

Moles Calculation

To grasp the concept of moles calculation thoroughly, think about it as finding out how many individual particles, like molecules or atoms, are present in a certain amount of substance. In chemistry, the number of moles can be calculated using the formula \( M = \frac{n}{V} \), where:\

• \( M \) represents molarity,
• \( n \) stands for the number of moles,
• \( V \) indicates the volume of solution in liters.

If you have a 0.755 M solution of cobalt(II) chloride (\( ext{CoCl}_2\)), and the solution's volume is \(0.0245\) liters, you multiply the molarity by the volume to find the number of moles:

• \( n = M \times V = 0.755 \times 0.0245 = 0.0184975 \text{ moles}\).

This calculation is crucial, as knowing the number of moles allows chemists to proceed with further calculations, such as finding the mass of the solute, which is our next step.

Molar Mass

Molar mass links the number of moles of a substance to its mass, making it a fundamental concept in chemistry. It's essential for translating the amount of a chemical you have in moles into grams, providing a tangible measure of how much "stuff" you're dealing with.

To calculate molar mass, you add up the atomic masses of each element in a compound, based on their proportions. For cobalt(II) chloride (\( ext{CoCl}_2\)), you have one atom of cobalt (Co) and two atoms of chlorine (Cl). With the atomic mass of Co being \(58.93\, \text{g/mol}\) and Cl \(35.45\, \text{g/mol}\), the molar mass is:

• \(58.93 + 2 \times 35.45 = 129.83 \text{ g/mol}\).

Once you know the molar mass, you can easily find the mass of a substance if you know the moles present.

• For example, for \(0.0184975\) moles of \( ext{CoCl}_2\), multiply the moles by the molar mass: \(0.0184975 \times 129.83 = 2.401 \text{ g}\).

This connection between moles and mass is critical for various chemical calculations, ensuring precise and reliable results.