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sierram298
16.10.2019 •
Chemistry
The boiling point of water is 100.0°c at 1 atmosphere. how many grams of sodium acetate (82.04 g/mol), must be dissolved in 283.0 grams of water to raise the boiling point by 0.450°c?
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Ответ:
I have to weight 10,04 g of acetate sodium
Explanation:
This is the colligative propertie about elevation of boiling point
ΔT = Kb . m . i
ΔT is the difference between T° at boiling point of the solution - T° at boiling point of the solvent pure - we have this data 0,450°C
Kb means ebulloscopic constant (0,52 °C.kg/m .- a known value for water)
m means molality (moles of solute in 1kg of solvent)
i means theVan 't Hoff factor (degree of dissociation for a compound)
For the sodium acetate is 2
NaCH3COO ---> Na+ + CH3COO-
0,450°C = 0,52°C.kg/m . m . 2
0,450°C / (0,52°C.kg/m . 2) = m
0,432 m/kg = m
This number means I have 0,432 moles of acetate sodium, my solute in 1kg of water, my solvent. But I don't have 1000 g (1kg) I only have 283 g so let's make the rule of three:
1000 g 0,432 moles
283 g (283g .0,432m) / 1000g = 0,122 moles
Now that I have the moles of acetate sodium, I have to find the mass.
Moles . molar mass = mass
0,122 moles . 82.04 g/mol = 10,04 g
Ответ:
A
Explanation: