Question

What volume of \(1.772 \mathrm{M} \mathrm{BaCl}_{2}\) is needed to obtain \(123 \mathrm{~g}\) of \(\mathrm{BaCl}_{2} ?\)

Short answer

333 milliliters of 1.772 M BaCl2 solution is needed.

Step-by-step solution

Calculate the Molar Weight of BaCl2

The molar weight of a compound is the sum of the molar weights of its constituent elements. For \(\text{BaCl}_2\): \(\text{Ba}\) has a molar mass of approximately 137.33 g/mol, and \(\text{Cl}\) has a molar mass of approximately 35.45 g/mol. Thus, the molar weight of \(\text{BaCl}_2\) is: \[ 137.33 + 2 \times 35.45 = 208.23 \text{ g/mol} \]

Convert Mass of BaCl2 to Moles

To find out how many moles are in 123 grams of \(\text{BaCl}_2\), use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] Plugging in the numbers: \[ \frac{123 \text{ g}}{208.23 \text{ g/mol}} \approx 0.5907 \text{ moles} \]

Use Molarity to Find Volume

Molarity (M) is defined as moles of solute per liter of solution. The given molarity is 1.772 M, meaning 1.772 moles per liter. Rearrange the molarity equation \(M = \frac{\text{moles}}{\text{volume}}\) to solve for volume: \[ \text{volume} = \frac{\text{moles}}{M} = \frac{0.5907 \text{ moles}}{1.772 \text{ M}} \approx 0.333 \text{ liters} \]

Convert Volume to Milliliters

To convert liters to milliliters, multiply by 1000 (since 1 liter = 1000 milliliters). \(0.333 \text{ liters} \times 1000 = 333 \text{ milliliters}\). So, you need 333 milliliters of 1.772 M \(\text{BaCl}_2\) solution.

Key concepts

Molar mass

Understanding molar mass is crucial for making calculations in chemistry. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It summarizes the mass contribution of each constituent element within a compound. For example, in barium chloride (\(\text{BaCl}_2\)) you'll add the atomic masses of barium (Ba) and chlorine (Cl) to find its molar mass.
- Barium (Ba) has an atomic mass of approximately 137.33 g/mol. - Chlorine (Cl) has an atomic mass of approximately 35.45 g/mol.
To determine the molar mass of \(\text{BaCl}_2\), sum these values, taking into account that there are two chlorine atoms, yielding:\[ 137.33 + 2 \times 35.45 = 208.23 \text{ g/mol} \]This value is now used in further calculations to convert between mass and moles.

Solution preparation

To prepare a solution of a specified concentration, understanding molarity is essential. Molarity is the number of moles of solute (the substance being dissolved) per liter of solution. In this scenario, when we say a solution of \(1.772 \text{ M } \text{BaCl}_2\), it signifies there are 1.772 moles of \(\text{BaCl}_2\) per liter of the mixture.

The calculation involves converting a given mass of solute to moles, using its molar mass. Once you know the number of moles, you can determine how much solution you need, using the molarity equation: \[ M = \frac{\text{moles of solute}}{\text{volume of solution}} \] Rearranging this, you can solve for the required volume:\[ \text{volume} = \frac{\text{moles}}{M} \] This equation is fundamental when mixing chemical solutions accurately in a laboratory or industrial setting.

Unit conversion

Unit conversion is an everyday part of chemistry, bridging different measurement systems for accurate calculations. Two major conversions within molarity computations are mass to moles and volume of solution to different unit measures.
- Conversion from grams to moles requires dividing the mass by the molar mass. For example, with \(123\text{ g}\) of \(\text{BaCl}_2\) and a molar mass of \(208.23 \text{ g/mol}\), calculate moles as:\[ \text{moles} = \frac{123 \text{ g}}{208.23 \text{ g/mol}} \approx 0.5907\text{ moles} \]
- Conversion from liters to milliliters involves multiplying by 1000 (since 1 liter equals 1000 milliliters). So, a calculated volume of \(0.333\text{ L}\) translates to:\[ 0.333 \text{ L} \times 1000 = 333 \text{ mL} \]
These conversions ensure consistent and accurate results in solution preparation and analysis. Understanding unit conversion makes it easier to interpret and solve real-world chemistry problems.