Question

Show one sample calculation to convert from % mass to molality. This is a ethylene glycol...

Show one sample calculation to convert from % mass to molality.

This is a ethylene glycol solution.

Mass % Ethylene glycol= 2.00, molality=0.32

mass %: 2= (0.50/25.01)x100

molality: 0.32=(0.50 x (1/62.06)) / ((25.01-0.50) x 1/1000)

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Mass%= Mass of solute -xloo mass of solute & Solvent) 0.5 2 = x100 25:0) Here mass of solate =0.59 Total mass of solute tsolv

Add a comment
Know the answer?
Add Answer to:
Show one sample calculation to convert from % mass to molality. This is a ethylene glycol...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Show one sample calculation to find the Tf for a 52.0% mass solution (again, use the...

    Show one sample calculation to find the Tf for a 52.0% mass solution (again, use the trendline from the molality graph). (Hint: convert 52.0% to correct units!) The equation is Tf=0.485(m)+-0.134 where Tf is freezing temperature and m is molality. I'm guessing you have to convert percent mass to molality but I don't know how to do that. EDIT: It is a Ethylene glycol solution, so molar mass is 62.06.

  • What is the molality of ethylene glycol, C2H4(OH)2, in a solution prepared from 2.22 × 103...

    What is the molality of ethylene glycol, C2H4(OH)2, in a solution prepared from 2.22 × 103 g of ethylene glycol and 2.00 × 103 g of water?

  • 15. What is the molality of a solution made from 1 gal of ethylene glycol (density...

    15. What is the molality of a solution made from 1 gal of ethylene glycol (density 1.11 g/ml) and I gal of water (density 1.00 g/ml)? 4 pts 16. Determine the freezing point and boiling point for the aqueous solution in number 15 above. Water is the solvent. 4 pts

  • A solution of ethylene glycol in water at 20 degrees celsius has a mass percent of...

    A solution of ethylene glycol in water at 20 degrees celsius has a mass percent of 9.78% of ethylene glycol with a density of 1.0108 g/mL. The freezing point depression constant for water (solvent for all solutions) is Kf=-1.86 percent celsius kg/mol and the boiling point elevation constant is Kb=0.512 degrees celsius kg/mol. The density of neat water at 20.0 degrees celsius is 0.9982 g/mL. Answer the following: 1. What is the molarity of the solution? 2. What is the...

  • A solution of ethylene glycol in water at 20.0°C has a mass percent of 8.25% of...

    A solution of ethylene glycol in water at 20.0°C has a mass percent of 8.25% of ethylene glycol with a density of 1.0087 g/mL. The freezing point depression constant for water (which you can assume is the solvent for all solutions) is K1.86°C kg/mol and the boiling point elevation constant is Kb the following: 0.512°C kg/mol. The density of neat water at 20.0°C is 0.9982 g/ml. Answer 1. What is the molarity of the solution? 2. What is the molality...

  • 2. Calculate the concentrations of solutions 1-7 from the procedure step 2. Show one sample calculation....

    2. Calculate the concentrations of solutions 1-7 from the procedure step 2. Show one sample calculation. Test Tube # Concentration (M) 2. Prepare one solution at a time by pipeting the following volumes into a small beaker for mixing, then transfer the solution to the appropriate test tube. Test Tube No. 0.150 M COCI, (m.) 2.00 4.00 5.00 6.00 7.00 8.00 10.00 Distilled H,O (mL) 8.00 6.00 5.00 4.00 3.00 2.00 0.00

  • i need help using the formula in the 3rd picture , to figure picture 2(the molar...

    i need help using the formula in the 3rd picture , to figure picture 2(the molar mass). I included needed information in picture 1 Pure TBA, T9 Ethylene Glycol/TBA Solution, T Calculation of Molar Mass of Ethylene Glycol Mass of ethylene glycol used (total amount in solution) .94 253 t Mass of TBA used Freezing point of pure TBA, T Freezing point of solution, T, oC oc oC Freezing point depression (total dgpression), 471 Total molal concentration m of ethylene...

  • 26.)Show the calculation of the mass of NaNO3 needed to make 450 ml of a 0.356...

    26.)Show the calculation of the mass of NaNO3 needed to make 450 ml of a 0.356 M solution. 25.)Show the calculation of the molarity of a solution made by dissolving 15.9 grams of Ca3(PO4)2 to make 350 ml of solution. 24.)Show the calculation of the molarity of a solution made by dissolving 12.6 grams of NaNO3 to make 200 ml of solution. 23.Show the calculation of the molality of a solution made by dissolving 15.9 grams of Ca3(PO4)2 in 400...

  • Calculation required for Q. 2: Calculation of mass of CaCO3 sample for analysis.

    calculation required for Q. 2: Calculation of mass of CaCO3 sample for analysis. solid calcium carbhotae samples, calcium bicarbonate, CalHCO), i abo prscn. 1. In some calciunm for its reaction with hydrochloric acid. (og) 2 Experimental Procedure, Part A.2a. Complete the calculation required to appear on the Report Sheet (here and Report Sheet). 3. a. Experimental Procedure, Part A.3b. Explain how a water-filled graduated cylinder is inverted in a pan of water. b. Experimental Procedure, Part A.4. What is the...

  • The molar mass of a compound expresses the ratio of mass to moles Part A molar...

    The molar mass of a compound expresses the ratio of mass to moles Part A molar mass- _mass in gr tert-Butyl alcohol is a solvent with a Kf of 9.10 °C/m and a freezing point of 25.5 °C. When 0.807 g of an unknown colorless liquid was dissolved in 11.6 g of tert-butyl alcohol, the solution froze at 15.3°C number of moles This quantity can be determined experimentally by accurately measuring the mass of the sample and determining the corresponding...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT