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

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Discussion Overview

The discussion revolves around the calculation of molarity and molality in a chemistry homework context, specifically addressing the definitions and application of these concepts in relation to a professor's website that presents potentially incorrect information.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses skepticism about the accuracy of a professor's website regarding molarity and molality calculations.
  • Another participant agrees with the initial calculations, suggesting that the professor's answer is likely incorrect.
  • A third participant points out that the website misdefines molality as moles of solute per kg of solution instead of per kg of solvent.
  • A fourth participant adds that the densities of various solutions listed on the website are also incorrect, although one density is noted as being close to accurate.

Areas of Agreement / Disagreement

Participants generally agree that there are inaccuracies in the professor's website, but there is no consensus on the implications of these inaccuracies or the correctness of the calculations presented.

Contextual Notes

There are unresolved assumptions regarding the definitions of molarity and molality, as well as the implications of the density values provided in the discussion.

Qube
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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.
 
Last edited by a moderator:
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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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