HomeworkLib
Question

6. A 1.07 mg sample of a compound was dissolved in 78.1 mg of camphor. The...

6. A 1.07 mg sample of a compound was dissolved in 78.1 mg of camphor. The resulting solution melted at 176.0 C. What is the molecular mass of the compound? The melting point of pure camphor is 179.5 C. The freezing-point-depression constant for camphor Kf is 40.0 C/m.

7. A solution is prepared by dissolving 5.88 g of an unknown nonelectrolyte in enough water to make 0.355 L of solution. The osmotic pressure of the solution is 1.21 atm at 27 C. What is the molar mass of the solute? (R = 0.08206 Latm/molK)

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:
6. A 1.07 mg sample of a compound was dissolved in 78.1 mg of camphor. The...
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

  • A 1.07 mg sample of a compound was dissolved in 78.1 mg of camphor, a solvent...

    A 1.07 mg sample of a compound was dissolved in 78.1 mg of camphor, a solvent that melts at 179.5 *C and has an unusually high Kfp of 40.0 *C/molal. The freezing point of the resulting solution dropped to 176.0 *C. a. Calculate delta t fp b. Calculate the molality of the solution c. Calculate the moles of solute in the solution d. Calculate the molar mass of the compound

  • (6 pt) 4 A 1.07-mg sample of a non-ionic compound was dissolved in 78.1 mg of...

    (6 pt) 4 A 1.07-mg sample of a non-ionic compound was dissolved in 78.1 mg of camphor, a solvent that melts at 179.5°C and has an unusually high Kr of 40.0 mola If the freezing point of the solution is 176.0°C: a) Calculate AT, then determine the molality (m) of this solution: b) Calculate the moles of solute in this solution: e) Calculate the molar mass ()of e coound:

  • A solution is prepared by dissolving 0.107 g of an unknown non-electrolyte in 88.1 g of...

    A solution is prepared by dissolving 0.107 g of an unknown non-electrolyte in 88.1 g of camphor. The freezing point of the solution was measured to be 178.0�C. What is the molecular weight of the compound? (The freezing point of pure camphor is 179.5�C and kf= 40�C/m)?

  • A 275 mg sample of a nonelectrolyte compound was dissolved in water to produce 10.0 mL...

    A 275 mg sample of a nonelectrolyte compound was dissolved in water to produce 10.0 mL of a solution at 25°C. The osmotic pressure of this solution was measured and found to be 3.67 atm. What is the molar mass of this compound? O 183 g/mol 18.3 g/mol 101 g/mol 1560 g/mol 1830 g/mol Question 25 (1 point) Which statement below is true? O Nonpolar molecules can dissolve in polar solvents. Gases are soluble only in nonpolar solvents. lonic compounds...

  • A sample of 4.6434 g of benzoic acid, C_7H_6O_2, is dissolved in 50.0033 g of camphor....

    A sample of 4.6434 g of benzoic acid, C_7H_6O_2, is dissolved in 50.0033 g of camphor. what is the freezing point of the resulting solution? For camphor, T_f = 179.5 degree C, K_f = 40.00 degree C/m.

  • A solution is made by dissolving 58.125g of sample of an unknown, nonelectrolyte compound in water....

    A solution is made by dissolving 58.125g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is exactly 750.0g. The boiling point of this solution is 100.220 ∘C. What is the molecular weight of the unknown compound? Solvent Formula Kf value* (°C/m) Normal freezing point (°C) Kb value (°C/m) Normal boiling point (°C) water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3...

  • A solution is made by dissolving 0.592 mol of nonelectrolyte solute in 767 g of benzene....

    A solution is made by dissolving 0.592 mol of nonelectrolyte solute in 767 g of benzene. Calculate the freezing point, Te, and boiling point, Tb, of the solution. Constants can be found in the table of colligative constants. T = Colligative Constants Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Solvent Formula Kf value* Normal freezing Kb value Normal boiling (°C/m) point (°C) (°C/m) point (°C) water H20 1.86 0.00 0.512 100.00 benzene 5.12 5.49 2.53 80.1...

  • 1.A solution that contains 15.581 grams of benzene (molar mass = 78.1 g/mol) in 420 g...

    1.A solution that contains 15.581 grams of benzene (molar mass = 78.1 g/mol) in 420 g of cyclohexane freezes at a temperature 9.50 ºC below the freezing point of pure cyclohexane. Calculate the molar freezing point depression constant, Kf, for cyclohexane. 2. The molar boiling point elevation constant for water is 0.512 ºC/m. What is the boiling point of a solution containing 673 g of water and 78.0 g of CaCl2? 3. A solution containing 32.0 g of an unknown...

  • A solution is made by dissolving 0.585 mol of nonelectrolyte solute in 877 g of benzene....

    A solution is made by dissolving 0.585 mol of nonelectrolyte solute in 877 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants can be found in the table of colligative constants. Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Solvent Formula Kf value* (°C/m) Normal freezing point (°C) Kb value (°C/m) Normal boiling point (°C) water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8...

  • When 1.81 g of Vitamin K is dissolved in 25.0 g of camphor, the freezing point...

    When 1.81 g of Vitamin K is dissolved in 25.0 g of camphor, the freezing point of the solution is lowered by 6.06 C. Look up the freezing paint and the Kf constant for camphor in the Colligative Constants table. Calculate the molar mass of Vitamin K.

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