Calculating Molarity and Molality: Tips and Tricks | Chemistry Homework Help

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SUMMARY

This discussion focuses on the accurate calculation of molarity and molality, highlighting the distinction between the two concepts. Molarity is defined as moles of solute per liter of solution, while molality is moles of solute per kilogram of solvent. The participant critiques a university professor's website for incorrectly defining molality and provides a detailed calculation involving sucrose, demonstrating that the molality derived from the provided data does not align with the expected results. Key densities for various solutions, including 0.840 M sucrose and 4.91 M NaOH, are also mentioned to emphasize the importance of accurate density values in these calculations.

PREREQUISITES
  • Understanding of molarity and molality definitions
  • Basic knowledge of solution density and its implications
  • Ability to perform mole calculations and conversions
  • Familiarity with the concept of solute and solvent in solutions
NEXT STEPS
  • Research the differences between molarity and molality in detail
  • Learn how to calculate density of solutions and its impact on molarity and molality
  • Explore the implications of solution concentration on chemical reactions
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Chemistry students, educators, and anyone involved in laboratory work requiring precise calculations of solution concentrations will benefit from this discussion.

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Homework Statement



Screenshot of some prof's website. This is just ridiculous.

http://i.minus.com/jnUa604Tz6bKX.png

Homework Equations



Molarity to molality.

Molality is moles / kg of solvent.

Molarity is moles / volume of solution.

The Attempt at a Solution



Pretty sure all the ones in the screenshot from the webpage of some university professor are wrong.

For 1)

a) Assume 1000 mL of solution. We therefore have 0.84 moles of sucrose (0.84 moles / 1 L = 0.84 M)

b) This means we have 1120 g of solution.

c) Solution mass = solute mass + solvent mass.

d) Solute mass = 0.84 moles * 342 g/moles = 287 grams.

e) 1120 - 287 = solvent mass = 833 g = 0.833 kg.

f) 0.84 / 0.833 isn't anywhere close to 3/4 molal solution.
 
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I see nothing wrong with your math, so I guess his answer is wrong.
 
The website incorrectly assumed that molality is the number of moles of solute per kg of solution (rather than the correct definition which you gave).
 
Not to mention the fact density of 0.840 M sucrose solution is 1.1079 g/mL, density of 4.91 M NaOH solution is 1.1840 g/mL, and of 0.79 M NaHCO3 solution is 1.0455 g/mL. Neither is listed correctly (although sucrose is close).
 

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