In the Jalloh Company lab, a 0.2828 m solution contains 49.117 grams of lithium bromide. What volume of water was added to the solute make this solution? Assume the density of water is 1.0000 g cm-3.


To determine the volume of water added to the solute to make this solution, we can follow these steps:

### Given Data:
1. **Molality (m):** 0.2828 m (molality is moles of solute per kilogram of solvent).
2. **Mass of solute (Lithium bromide, LiBr):** 49.117 g.
3. **Density of water:** 1.0000 g/cm³.
4. **Molar mass of LiBr:** Li (6.941 g/mol) + Br (79.904 g/mol) = 86.845 g/mol.

### Step 1: Calculate moles of LiBr.
The number of moles of lithium bromide is calculated using the formula:

\[
\text{Moles of LiBr} = \frac{\text{Mass of LiBr}}{\text{Molar mass of LiBr}} = \frac{49.117 \, \text{g}}{86.845 \, \text{g/mol}} = 0.5655 \, \text{mol}
\]

### Step 2: Use the definition of molality.
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent (water in this case). The formula for molality is:

\[
m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}
\]

Rearranging for the mass of solvent:

\[
\text{Mass of solvent} = \frac{\text{Moles of solute}}{m}
\]

Substitute the given molality and the moles of solute:

\[
\text{Mass of solvent} = \frac{0.5655 \, \text{mol}}{0.2828 \, \text{mol/kg}} = 1.999 \, \text{kg}
\]

### Step 3: Convert the mass of water to volume.
Since the density of water is 1.0000 g/cm³, which is equivalent to 1.0000 g/mL, we can convert the mass of water to volume. Recall that 1 kg of water is 1000 g.

\[
\text{Volume of water} = \frac{1.999 \, \text{kg} \times 1000 \, \text{g/kg}}{1.0000 \, \text{g/mL}} = 1999 \, \text{mL} = 1.999 \, \text{L}
\]

### Conclusion:
The volume of water added to make the solution is **1.999 liters**.

Would you like further details or have any questions?

Here are 5 related questions to explore:
1. How does molality differ from molarity in solution concentration?
2. What effect would temperature changes have on the volume of the water added?
3. How would the calculation change if the density of water were not 1.000 g/cm³?
4. What are the potential errors in assuming a constant density of water for this calculation?
5. How would the result differ if a different solute, like sodium chloride, were used?

**Tip:** When converting between mass and volume, always check the units of density to avoid errors in dimensional analysis.

In the Jalloh Company lab, a 0.2828 m solution contains 49.117 grams of lithium bromide. What volume of water was added to the solute make this solution? Assume the density of water is 1.0000 g cm-3.

Learn how to determine the volume of water added to a 0.2828 m solution of lithium bromide containing 49.117 grams of solute. This problem involves key concepts like molality, molar mass, and mass-to-volume conversion. Perfect for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation