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The boiling point of water is 100.0°C at 1 atmosphere. (Kb(water) = 0.512°C/m)    How many...

The boiling point of water is 100.0°C at 1 atmosphere. (Kb(water) = 0.512°C/m)   

How many grams of calcium iodide (293.9 g/mol), must be dissolved in 280.0 grams of water to raise the boiling point by 0.300°C ?

g calcium iodide.

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Answer #1

delta Tb = i * Kb * molality

0.300 = 3 * 0.512 * molality

0.300 = 1.54 * molality

molality = 0.195 m

molality = number of moles / mass of solution in kg

0.195 = number of moles / 0.280 kg

number of moles of calcium iodide = 0.195 * 0.280 = 0.0546 mole

number of moles of calcium iodide = mass of calcium iodide / molar mass of calcium iodide

0.0546 mole = mass of calcium iodide / 293.887 g/mol

mass of calcium iodide = 0.0546 mole * 293.887 g/mol = 16.0 g

Therefore, the mass of calcium iodide = 16.0 g

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