Question

2. I. M. Fumblefingers determined the mass of a 100 mL volumetric flask but did not notice that it contained about 1 mL of wa
0 0
Add a comment Improve this question Transcribed image text
Answer #1

By definition 8% solution means we have to

weigh 8 grams of solute --> then we have to carefully transfer it into a 100 mL volumetric flask --> fill the flask up to the mark with pure distilled water (solvent)

this gives a perfect 8% solution.

Now addressing questions:

a) The concentration of the solution (solute + solvent) be no different from the correct value. Because we have weighed exactly 8 grams of sugar and transferred it into 100 mL volumetric flask then anyway we have to add water, so even 1 mL of higher amount could not affect our concentration, unless we have added extra amount of water which crosses the mark or weighed more than 8 grams of sugar.

b) Since we have followed exact protocol to prepare 8% solution, so it would not affect calculated density also, so the density fumblefingers determines for this solution will be no different from the correct value, unless we have added extra amount of water which crosses the mark or weighed more than 8 grams of sugar.

Hope this helped you!

Thank You So Much! Please Rate this answer as you wish.("Thumbs Up")

Add a comment
Know the answer?
Add Answer to:
2. I. M. Fumblefingers determined the mass of a 100 mL volumetric flask but did not...
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
  • A penny having a mass of 2.5340 g was dissolved in 20 mL of 8 M HNO3. The resulting solution was transferred to a 100.00-ml volumetric flask and diluted to the mark with water.

    A penny having a mass of 2.5340 g was dissolved in 20 mL of 8 M HNO3. The resulting solution was transferred to a 100.00-ml volumetric flask and diluted to the mark with water. Then, a sample solution was prepared by transferring 10.00 mL of this solution to a 25.00- ml volumetric flask, adding 2.00 mL of 15 M NH3 and diluting to the mark with waterThe concentration of [Cu(NH3)4]2+ in the sample solution (determined using a Spectro Vis spectrophotometer)...

  • Question 1 2 pts 3.34 mL of an unknown glucose solution was diluted in a 100.00 mL of volumetric flask with distill...

    Question 1 2 pts 3.34 mL of an unknown glucose solution was diluted in a 100.00 mL of volumetric flask with distilled water. 10mL of that solution and 4.00 mL of the ferricyanide reagent was placed in a second 100.00 mL volumetric flask and diluted to the mark with distilled water. Spectrophotometric analysis of the glucose/ferricyanide solution showed that the concentration of glucose was 0.070 M. What is the concentration of glucose in the unknown solution?

  • Question 1 2 pts 3.34 mL of an unknown glucose solution was diluted in a 100.00 mL of volumetric flask with distill...

    Question 1 2 pts 3.34 mL of an unknown glucose solution was diluted in a 100.00 mL of volumetric flask with distilled water. 10mL of that solution and 4.00 mL of the ferricyanide reagent was placed in a second 100.00 mL volumetric flask and diluted to the mark with distilled water. Spectrophotometric analysis of the glucose/ferricyanide solution showed that the concentration of glucose was 0.070 M. What is the concentration of glucose in the unknown solution?

  • You diluted a 12.96 ml vinegar sample into a 100 ml volumetric flask with distilled water....

    You diluted a 12.96 ml vinegar sample into a 100 ml volumetric flask with distilled water. You titrated a 25.00 mL aliquot of that solution and determined the concentration of acetic acid to 0.048 M. What was the concentration of acetic in the original sample.

  • To a 25.00 mL volumetric flask, a lab technician adds a 0.300 g sample of a...

    To a 25.00 mL volumetric flask, a lab technician adds a 0.300 g sample of a weak monoprotic acid, HA, and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0867 M KOH. She reaches the endpoint after adding 41.23 mL of the KOH solution. Determine the number of moles of the weak acid in the solution. moles of weak acid: mol Determine the molar mass of the weak acid. molar mass =...

  • To a 25.00 mL volumetric flask, a lab technician adds a 0.350 g sample of a...

    To a 25.00 mL volumetric flask, a lab technician adds a 0.350 g sample of a weak monoprotic acid, HA, and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0868 M KOH. She reaches the endpoint after adding 43.65 mL of the KOH solution. Determine the number of moles of the weak acid in the solution. moles of weak acid: mol Determine the molar mass of the weak acid. molar mass =...

  • To a 25.00 mL volumetric flask, a lab technician adds a 0.375 g sample of a...

    To a 25.00 mL volumetric flask, a lab technician adds a 0.375 g sample of a weak monoprotic acid, HA , and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0923 M KOH . She reaches the endpoint after adding 44.93 mL of the KOH solution. After the technician adds 15.35 mL of the KOH solution, the pH of the mixture is 5.73. Determine the p?a of the weak acid.

  • D Question 1 2 pts 0.635g of NaOH was placed in a 250 mL volumetric flask...

    D Question 1 2 pts 0.635g of NaOH was placed in a 250 mL volumetric flask and distilled water was added to the 250.00 mL mark. What volume of 0.231M sulfuric acid, in mL, is required to titrate a 25.00 mL aliquot of the NaOH solution? Round your answer to two decimal places. Molar Mass: Na 23.0 g/mol, S 32.1 g/mol, O 16.0 g/mol, H 1.0g/mol

  • Pour 2.5 mL of the 1M glucose (C6H12O6) solution from the flask into the 100 mL...

    Pour 2.5 mL of the 1M glucose (C6H12O6) solution from the flask into the 100 mL volumetric flask. Use the “dilution equation” to determine the volume in mL for the required 0.025M solution, given that the original solution is 1M and you used 2.5 mL Fill the volumetric flask to the 100 mL mark with distilled water. Note: The flask already contains a volume of 2.5 mL. 1. Compute the volume of the diluted 0.025 M solution.   2. What volume...

  • A student weighs out 14.6 g of AlBrz, transfers it to a 300 mL volumetric flask,...

    A student weighs out 14.6 g of AlBrz, transfers it to a 300 mL volumetric flask, adds enough water to dissolve the solid and then adds water to the 300 mL mark on the neck of the flask. Calculate the concentration (in molarity units) of aluminum bromide in the resulting solution? Calculate the mass, in grams of sodium iodide that must be added to a 250 mL volumetric flask in order to prepare 250 mL of a 0.138 M aqueous...

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