ZuoTi.Pro
Question

This solution of unknown a in napthalene was prepared by adding 2.70 mL of unknown a...

This solution of unknown a in napthalene was prepared by adding 2.70 mL of unknown a (D=1.25g/mL) to 19.950 g of napthalene. using this data and your answer to the previous question (my answer was freezing point depression=8 degrees C). calculate the molar mass of unknown a. kf of napthalene is 6.94 c*kg/mol

please show all work
science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

0 0
Add a comment
Know the answer?
Add Answer to:
This solution of unknown a in napthalene was prepared by adding 2.70 mL of unknown a...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

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

  • An unknown sample weighs 0.2436 g was dissolved in 20.0 mL of cyclohexane (density = 0.779...

    An unknown sample weighs 0.2436 g was dissolved in 20.0 mL of cyclohexane (density = 0.779 g/mL). The freezing point depression (∆Tf) was 2.5°C. Calculate the molar mass of the unknown substance. (Kf = 20.5 °C/m)     ∆Tf = Kf m where Kf is the freezing point depression constant and m is molality. 526 g/mol 1.50 g/mol 128 g/mol 2.44 g/mol

  • A sugar solution is prepared by adding 3.50 g of sugar to 22.5 mL of a...

    A sugar solution is prepared by adding 3.50 g of sugar to 22.5 mL of a pure liquid and the freezing point of the sugar solution was determined to be -99.84 C. If the normal freezing point of the pure liquid is -97 C and the density is .792g/mL, what is the freezing point depression constant in degrees celsius?

  • The molar mass of unknown molecular compound is determined using freezing point depression. The substance is...

    The molar mass of unknown molecular compound is determined using freezing point depression. The substance is dissolved in benzene (C6H), an organic solvent. Use the data below to determine the molar mass of the substance. 8.65 g Mass of unknown compound Freezing point of solution Volume of benzene -10.21°C 35.0 mL 5.53°C Freezing point of benzene Density of benzene Kf for benzene 0.8765 g/mL 5.12°C/m Molar mass = g/mol

  • A solution is prepared by dissolving 2.00 g of glycerin (C3H8O3) in 201 g of ethanol...

    A solution is prepared by dissolving 2.00 g of glycerin (C3H8O3) in 201 g of ethanol (C2H5OH). The freezing point of the solution is _______ degrees celsius. The freezing point of pure ethanol is -114.6 degrees celsius at 1atm. The molal-freezing point depression constant (Kf) for ethanol is 1.99 degrees celsius/m. The molar masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol, respectively.

  • A solution is prepared by dissolving 3.04 g of an unknown nonelectrolyte in 16.00 g of...

    A solution is prepared by dissolving 3.04 g of an unknown nonelectrolyte in 16.00 g of cylcohexane (tf = 6.6 oC and kf = 20.0 oC/m). What is the molar mass of the unknown compound if the solution has a freezing point of -6.7 oC? Please see https://youtu.be/lwWZ5hdB61M for assistance. Mm = g/mol

  • Need help with freezing point depression expeiment volume of acetic acid 30 volume of unknown solute...

    Need help with freezing point depression expeiment volume of acetic acid 30 volume of unknown solute in solution1    1.5 volume of unknown solute in solution 2 1.5 freezing point of pure acetic acid 16.3 freezing point of solution 1 14.5 freezing point of solution 2 11.3 average molar mass of unknown solute Calculations for Part 1 The density of acetic acid is 1.049 g/mL and Kf(acetic acid) =3.90 °C·kg/mol The density of the unknown is 0.791 g/mL. 1.Calculate the masses...

  • 2. A solution is prepared by dissolving 7.00 g of glycerin (C3H8O3) in 201 g of...

    2. A solution is prepared by dissolving 7.00 g of glycerin (C3H8O3) in 201 g of ethanol (C2H5OH). The freezing point of the solution is __________°C. The freezing point of pure ethanol is -114.6 °C at 1 atm. The molal-freezing-point-depression constant (Kf) for ethanol is 1.88 °C/m. The molar masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol, respectively.

  • When 12.6-grams of an unknown compound is placed in 0.116-kg of benzene, the freezing point is...

    When 12.6-grams of an unknown compound is placed in 0.116-kg of benzene, the freezing point is changed by 2.34°C. The value of Kf= 4.90°C/m for benzene. Determine the molar mass of the unknown compound. Answer to 1 decimal place and include your units. Answer: Determine the molality of a solution when 32.3 grams of ethanol, C2H5OH (Molar mass = 46.08 g/mol) is dissolved in 0.947 kg of water. Answer to 2 decimal places. Answer: Find the freezing point of a...

  • Hi, I need some help with Chemistry. Q1: Q2: Here is some background for the questions: Thank y...

    Hi, I need some help with Chemistry. Q1: Q2: Here is some background for the questions: Thank you in advance. You add 0.0336 moles of benzoic acid solute to 16.00 g of an unknown solvent, which lowers the freezing point of the solvent by 8.6 °C. Calculate the freezing point depression constant (Kf) of the unknown solvent. You dissolve 1.00 g sample of an unknown solute is in 8.00 g of lauric acid, which lowers the freezing point by 5.0...

  • A student is determining the molar mass of an unknown compound based on the following data:...

    A student is determining the molar mass of an unknown compound based on the following data:   mass compound: 4.80 g             mass solvent (water): 0.022 kg             new freezing point: -2.50 oC             freezing point depression constant (Kf): 1.86 oC/molal (Freezing point of pure water = 0 oC) 0.0786 g/mol 162.33 g/mol 0.00616 g/mol 293.26 g/mol

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