A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 2 years ago

A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

  • This Question is Open
  1. Ryaan
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Calculate moles first: moles glucose = 15.5 g/ atomic mass = 15.5 g/ (6*12+12*1+6*16) = 15.5/180 = 0.086 Calculate molality next: (molality and molarity are the same when water is the solvent) 0.086 moles/ .245 kg water = 0.3515 m Calculate temp depression: -1.86 oC/M *.3515 m = 0.65 oC so the freezing point will drop from 0oC to -0.65oC.

  2. brianna0905
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i was so lost, but i get it more know

  3. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.