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Calculate the boiling point of a solution of 400.0 g of ethylene glycol (C₂H₆O₂) dissolved in 400.0 g of water. Kf = 1.86°C/m and Kb = 0.512°C/m. Use 100°C as the boiling point of pure water. Note: (Report your answer to the nearest whole number without decimal. In the answer box, only type the number. If answer is negative, place a minus sign in front of the number)

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Final answer:

The boiling point of a solution of 400 g ethylene glycol in 400 g water is calculated to be 108°C, after determining molality and applying the boiling point elevation formula using the given constants.

Step-by-step explanation:

The boiling point elevation of a solution can be calculated using the formula ΔTb = i * Kb * m, where ΔTb is the boiling point elevation, i is the van't Hoff factor (which is 1 for non-electrolytes like ethylene glycol), Kb is the ebullioscopic constant of the solvent (water in this case), and m is the molality of the solution.

First, calculate the moles of ethylene glycol (EG) by using its molar mass. The molar mass of ethylene glycol (C₂H₆O₂) is 62.07 g/mol. Therefore, the moles of ethylene glycol are 400.0 g / 62.07 g/mol = 6.44 mol.

Next, calculate the molality (m) of the solution: molality = moles of solute / kg of solvent = 6.44 mol / 0.400 kg = 16.1 m.

Now use the ebullioscopic constant of water (Kb) to find the boiling point elevation: ΔTb = 1 * 0.512°C/m * 16.1 m = 8.24°C. So, the boiling point of the solution is the boiling point of pure water plus the elevation: 100°C + 8.24°C = 108.24°C.

Rounded to the nearest whole number, the boiling point is 108°C.

User Nurealam Siddiq
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