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

[Qube]

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.

 

[Lamebert]

I see nothing wrong with your math, so I guess his answer is wrong.

 

[Chestermiller]

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).

 

[Borek]

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).

 

[DeeNos]

I would like to address the concerns raised by this content. While it is true that calculating molarity and molality can be a bit tricky, it is important to follow proper equations and methods to ensure accurate results. In this case, the calculations provided by the professor's website are incorrect.

For the first problem, the assumption of 1000 mL of solution is correct, but the calculation of 0.84 moles of sucrose is not. This should be 0.84 moles / 1 L = 0.84 M, as correctly stated in the problem. Additionally, the calculation for the mass of the solution should be 1000 mL * 1.12 g/mL = 1120 g, not 1120 mL.

Furthermore, the method used to calculate the mass of the solvent is incorrect. Instead, we should use the formula mass = moles * molar mass, which gives us 0.84 moles * 342 g/mol = 287 g. This means that the mass of the solvent is 1120 g - 287 g = 833 g, not 730 g as stated on the website.

Lastly, the molality of the solution should be calculated as 0.84 moles / 0.833 kg = 1.01 molal, not 3/4 molal as stated on the website.

It is important to double check calculations and use proper equations when solving problems in chemistry. This ensures accurate results and a better understanding of the concepts being studied. I would recommend seeking clarification from the professor or consulting a reliable source for correct calculations in the future.