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Calculate The Boiling Point Of A Solution Containing 0.456

Boiling Point Elevation Formula:

\[ T_b = T_0 + i K_b \left( \frac{0.456 / M}{W} \right) \]

°C
°C·kg/mol
g/mol
kg

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1. What is Boiling Point Elevation?

Boiling point elevation is a colligative property that occurs when a non-volatile solute is added to a solvent, raising its boiling point. The extent of elevation depends on the concentration of solute particles in the solution.

2. How Does the Calculator Work?

The calculator uses the boiling point elevation formula:

\[ T_b = T_0 + i K_b \left( \frac{0.456 / M}{W} \right) \]

Where:

Explanation: The formula calculates how much the boiling point increases when 0.456 g of solute is dissolved in a given mass of solvent.

3. Importance of Boiling Point Calculation

Details: Understanding boiling point elevation is crucial in various applications including chemical engineering, pharmaceutical formulations, food processing, and determining molecular weights of unknown compounds.

4. Using the Calculator

Tips: Enter the pure solvent boiling point, van't Hoff factor, ebullioscopic constant, molar mass of solute, and mass of solvent. Ensure all values are positive and appropriate units are used.

5. Frequently Asked Questions (FAQ)

Q1: What is the van't Hoff factor?
A: The van't Hoff factor (i) represents the number of particles a solute dissociates into in solution. For non-electrolytes, i = 1; for electrolytes, it depends on the degree of dissociation.

Q2: How do I find the ebullioscopic constant?
A: K_b is a solvent-specific constant. For water it's 0.512 °C·kg/mol, for benzene it's 2.53 °C·kg/mol, and for ethanol it's 1.22 °C·kg/mol.

Q3: Why is the solute mass fixed at 0.456 g?
A: This calculator is specifically designed for solutions containing exactly 0.456 g of solute, which may represent a standard experimental condition.

Q4: Can this be used for any solvent?
A: Yes, as long as you have the correct ebullioscopic constant (K_b) for that particular solvent.

Q5: What if the solute doesn't dissolve completely?
A: The calculation assumes complete dissolution. For partially soluble solutes, the actual concentration would be lower than calculated.

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