Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

$m_{SA}=10.24gSA$

Explanation:

Hello,

In this case, by studying the water's boiling point elevation, it is possible to determine the grams of sodium acetate needed to rise its boiling point by 0.450 ºC as shown below:

$\Delta T=Kb*m*i$

Because the van't Hoff factor of sodium acetate is 2. $Kb$ accounts for the waters boiling point elevation constant and $m$ for the sodium acetate's molality which leads to the calculation of the mass, in such a way, it results:

$m=\frac{\Delta T}{Kb*i}=\frac{0.450^0C}{0.51^0C/m*2}=0.441m=0.441\frac{molSA}{kgH_2O}$

Thus, the mass of sodium acetate (SA) turns out:

$m_{SA}=0.441\frac{molSA}{kgH_2O}*\frac{82.04gSA}{1molSA}*0.283kgH_2O\\m_{SA}=10.24gSA$

Best regards.