ZuoTi.Pro
Question

2. If your unknown solute sample weighed 0.634 grams, but some of it stuck to the test tube wall when you were pouring it in and did not dissolve in the t-buty alcohol, what sort of error will this cause in the calculated molar mass of your unknownsolute? Explain.
science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

the results will be wayvard as the amount of solute wil be less than the actually present in the test tube or the moles of sample dissolved will be less than expected ... as a result the molecular mass of the unknowm won't be accurate ... (that is the reason we take readings twice or thrice so as to see if some error has occured while performing the procedure)

0 0
Add a comment
Know the answer?
Add Answer to:
If your unknown solute sample weighed 0.634 grams, but some of it stuck to the test...
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

  • Chemistry: Molar mass determination and freezing point depression. If your unknown solute sample weighed 0.634 grams,...

    Chemistry: Molar mass determination and freezing point depression. If your unknown solute sample weighed 0.634 grams, but some of it stuck to the test tube wall when you were pouring it in and did not dissolve in the t-butyl alcohol, what sort of error will this cause in the calculated molar mass of your unknown solute? Explain. What would be the freezing point of an aqueous solution that contains 10.3 grams of ethylene glycol [(C_2H_4(OH)_2] in 100 mL of water?

  • Freezing Point Depression: Determination of the Molar Mass of an Unknown Substance Laboratory Record Unknown Identifier...

    Freezing Point Depression: Determination of the Molar Mass of an Unknown Substance Laboratory Record Unknown Identifier #4 (All unknowns are nonelectrolytes, soi-1) Determination of the freezing point of pure l-butyl alcohol: (Record time-temperature data on the next page) 29.62 1. Mass of empty test tube 2. Mass of test tube and t-butyl alcohol after determination of freezing point of pure t-butyl alcohol 50.25 20.71 3. Mass of t-butyl alcohol Determination of the molar mass of an unknown: (Record time-temperature data...

  • for question 1 the molar mass of unknown is 102g/mol hation by Freezing Point Depression Vernier...

    for question 1 the molar mass of unknown is 102g/mol hation by Freezing Point Depression Vernier Logger Pro for Windows Section Post-Laboratory Questions (Use the space provided and additional paper if necessary. Record your nemerical answers in the boxes provided.) 1. Obtain the correct molar mass of your unknown from your laboratory instructor. Calculate the percent error in the molar mass that you found. Torg/mol percent error, 2. A student determined the freezing point of pure t-butyl alcohol following the...

  • Calculate the molar mass of your unknown solute using the accepted literature Kf value of 8.3...

    Calculate the molar mass of your unknown solute using the accepted literature Kf value of 8.3 for t-butanol and the change in temperature of 9.10 degrees celcius. Grams of the unknown solute= 0.631g. Grams of solvent (t. butanol)= 12.230g.

  • 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...

  • Date Name Lab Section Partner(s) Determination of Molar Mass by Freezing Point Depression Data: Mass of...

    Date Name Lab Section Partner(s) Determination of Molar Mass by Freezing Point Depression Data: Mass of empty test tube 2.2.6! Mass of test tube plus approximately 10.0 g of benzophenone 33.39 The difference is the mass of benzophenone (solvent) 8 kg Mass of unknown (-1.0 g) 1:09: 48.50°c 42.08. Freezing point of pure benzophenone Freezing point of benzophenone containing the unknown The difference is A fp CO Calculations: (Show set-up with cancellation of units) 1. molality of solution 2. moles...

  • Data and Results: Calorimetry A. Specific Heat: Unknown metal number Unknown solid number Trial 2...

    Data and Results: Calorimetry A. Specific Heat: Unknown metal number Unknown solid number Trial 2 Trial 1 42.SI3 Mass of test tube and stopper Mass of stoppered test tube and metal Mass of calorimeter Mass of calorimeter and water Mass of water Mass of metal Initial temperature of water in calorimeter Initial temperature of metal Equilibrium temperature of metal and water in calorimeter 38.394g 53 13 g 63 to2g 0o C water (inal-onitiat oC oC Imetal qwater Specific heat of...

  • You are given a 0.553 gram sample of an unknown metal carbonate. You add some of...

    You are given a 0.553 gram sample of an unknown metal carbonate. You add some of the sample to a 10.0 mL sample of 0.255 M HNO3. When you have added exactly enough sample to react with all the acid, you reweigh your remaining sample and find its mass is 0.364 grams. (a). How many moles of acid reacted with the sample (b). How many grams of sample reacted with the acid? (c). How many moles of the unknown carbonate...

  • did i solve 1 a corectly or do my numbers need to be switched? Test Tube...

    did i solve 1 a corectly or do my numbers need to be switched? Test Tube M y 315 dilu LaiculdLIUIS PARTS B AND C Measuring the Freezing Points of Water and Aqueous Solutions 142 Laboratory Data for Determining Pring Pet T hree Decent Solutions Mass of Delonised Water salute Mass of Solute Freezing Point None 4.28 -1.84331 Naci 4, 4337 0.2380 -2.23293 Unknown A -0.2393 2.62978 Unknown B 4.04866 0.7395 -2.14573 1 Use the following steps to determine the...

  • 2) A random sample of black bears were weighed at various times of the year. Some...

    2) A random sample of black bears were weighed at various times of the year. Some of the bears were weighed in the spring, some in the summer and some in the fall. The bears were tagged so that the same bear was not measured more than once. Use a 1% significance level and the following Statcato statistics, graphs and ANOVA printout to test the population claim that the time of year (season) is related to the weight of the...

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