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At − 13.0 ∘ C −13.0 ∘C , a common temperature for household freezers, what is the maximum mass of aspartame (C14H18N2O5) you can add to 2.50 kg 2.50 kg of pure water and still have the solution freeze? Assume that aspartame is a molecular solid and does not ionize when it dissolves in water.

Respuesta :

Answer : The maximum mass of aspartame is, 5142.3 grams

Explanation :  Given,

Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]

Mass of water (solvent) = 2.50 kg

Molar mass of aspartame = 294.3 g/mole

Formula used :  

[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of aspartame}}{\text{Molar mass of aspartame}\times \text{Mass of water in Kg}}[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]\Delta T_s[/tex] = freezing point of solution = [tex]-13.0^oC[/tex]

[tex]\Delta T^o[/tex] = freezing point of water = [tex]0^oC[/tex]

i = Van't Hoff factor = 1 (for non-electrolyte)

[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]

m = molality

Now put all the given values in this formula, we get

[tex]0^oC-(-13.0^oC)=1\times (1.86^oC/m)\times \frac{\text{Mass of aspartame}}{294.3g/mol\times 2.50kg}[/tex]

[tex]\text{Mass of aspartame}=5142.3g[/tex]

Therefore, the maximum mass of aspartame is, 5142.3 grams

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