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?

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

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