Chemistry = Freezing Point & Molality

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SUMMARY

The discussion centers on calculating the percent by mass of sugar in a mixture of sugar (C12H22O11) and table salt (NaCl) dissolved in water, resulting in a freezing point depression of -2.24°C. The freezing point depression equation, delta T = (i)(Kf solvent)(m), is applied, where the van 't Hoff factor (i) for NaCl is 2 and the cryoscopic constant (Kf) for water is 1.86°C kg/mol. The initial calculation yielded a mass percent of sugar at 49.2%, while the correct answer is 53.8%, indicating a miscalculation in the molality or the total mass of the solute.

PREREQUISITES
  • Understanding of colligative properties, specifically freezing point depression
  • Familiarity with the van 't Hoff factor in solutions
  • Knowledge of molality calculations
  • Basic proficiency in mass percent calculations
NEXT STEPS
  • Review the concept of freezing point depression and its applications in chemistry
  • Learn about the van 't Hoff factor and its significance in colligative properties
  • Practice calculating molality with various solutes and solvents
  • Explore common mistakes in mass percent calculations and how to avoid them
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Chemistry students, educators, and anyone involved in solving problems related to colligative properties and solution chemistry.

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Homework Statement



You have a 10.4g mixture of sugar (C12H22O11) and table salt (NaCl). When this mixture is dissolved in 150g of water the freezing point is found to be –2.24 degrees centigrade. Calculate the percent by mass of sugar in the original mixture.

Homework Equations



delta T = (i) (Kf solvent)(m)
Mass % = mass of substance 1 / total mass of all substances

The Attempt at a Solution



First I tried to find the molality of the salt
delta T = 2.24 = 2 (1.86) (m)
m=.602151= mols salt/kg water

so .602151m = X mols salt/.15 kg water

X mols of salt = .090323
grams of salt = 5.27872 grams of salt.

So...grams of sugar in the mixture would be 5.12128 grams sugar.

The percent mass by sugar is then 5.12128/10.40...but that's 49.2% and our teacher (the exam key) said that the answer was 53.8%.

Does anyone know what I am doing wrong? Thanks so much! =)
 
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