Fill in the blanks. ltbRgt a. The mass of `MgCL_(2)` should be dissolved in `750 g` of water in order to prepare a `1.05 m` solution is……
b. The percentage composition (by mass) and mole fraction of each component is sugar containing `1000 g` of sugar in `2000 g` of water is.........

Explanation

Let's solve the question step by step.

### Part A: Mass of MgCl₂ for a 1.05 m solution

1. **Understanding Molality**:
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.
- The formula for molality is:
\[ m = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}} \]

2. **Given Information**:
- Molality (m) = 1.05
- Mass of solvent (water) = 750 g = 0.750 kg (since 1 kg = 1000 g)
- Molar mass of MgCl₂ = 95 g/mol (calculated from atomic masses: Mg = 24.31, Cl = 35.45)

3. **Rearranging the Formula**:
- We need to find the mass of solute (MgCl₂). Rearranging the formula gives us:
\[ \text{mass of solute} = m \times \text{molar mass of solute} \times \text{mass of solvent (kg)} \]

4. **Substituting the Values**:
- Substitute the values into the rearranged formula:
\[ \text{mass of MgCl₂} = 1.05 \times 95 \, \text{g/mol} \times 0.750 \, \text{kg} \]

5. **Calculating**:
- Calculate the mass of MgCl₂:
\[ \text{mass of MgCl₂} = 1.05 \times 95 \times 0.750 = 74.85 \, \text{g} \]

### Answer for Part A:
The mass of MgCl₂ that should be dissolved in 750 g of water to prepare a 1.05 m solution is **74.85 g**.

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### Part B: Percentage Composition and Mole Fraction of Sugar Solution

1. **Given Information**:
- Mass of sugar = 1000 g
- Mass of water = 2000 g
- Total mass of solution = Mass of sugar + Mass of water = 1000 g + 2000 g = 3000 g

2. **Calculating Percentage Composition**:
- **Percentage of Sugar**:
\[ \text{Percentage of Sugar} = \left(\frac{\text{mass of sugar}}{\text{mass of solution}}\right) \times 100 = \left(\frac{1000}{3000}\right) \times 100 = 33.33\% \]
- **Percentage of Water**:
\[ \text{Percentage of Water} = \left(\frac{\text{mass of water}}{\text{mass of solution}}\right) \times 100 = \left(\frac{2000}{3000}\right) \times 100 = 66.67\% \]

3. **Calculating Moles**:
- **Molar Mass of Sugar (C₁₂H₂₂O₁₁)**:
- Molar mass = \(12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \, \text{g/mol}\)
- **Moles of Sugar**:
\[ \text{Moles of Sugar} = \frac{\text{mass of sugar}}{\text{molar mass of sugar}} = \frac{1000 \, \text{g}}{342 \, \text{g/mol}} \approx 2.92 \, \text{moles} \]
- **Moles of Water**:
- Molar mass of water (H₂O) = 18 g/mol
\[ \text{Moles of Water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{2000 \, \text{g}}{18 \, \text{g/mol}} \approx 111.11 \, \text{moles} \]

4. **Calculating Mole Fractions**:
- **Mole Fraction of Sugar**:
\[ \text{Mole Fraction of Sugar} = \frac{\text{moles of sugar}}{\text{moles of sugar} + \text{moles of water}} = \frac{2.92}{2.92 + 111.11} \approx 0.026 \]
- **Mole Fraction of Water**:
\[ \text{Mole Fraction of Water} = 1 - \text{Mole Fraction of Sugar} = 1 - 0.026 \approx 0.974 \]

### Answer for Part B:
- Percentage composition by mass:
- Sugar: **33.33%**
- Water: **66.67%**
- Mole fractions:
- Sugar: **0.026**
- Water: **0.974**

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