The distribution coefficient between methylene chloride and water forsolute Y is 11. An amount of 50.0 g of Y is dissolved in 250 mL of water.a) What weight of Y would be removed from water with a single extractionwith 250-mL of methylene chloride? Show calculations and report to 1decimal place.

Given Information:Distribution coefficient (Kd) = 11Initial amount of solute Y in water = 50.0…

Answered: The distribution coefficient between…

Macroscale and Microscale Organic Experiments

7th Edition

ISBN:9781305577190

Author:Kenneth L. Williamson, Katherine M. Masters

Publisher:Kenneth L. Williamson, Katherine M. Masters

Chapter4: Recrystallization

Section: Chapter Questions

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The distribution coefficient between methylene chloride and water for solute Y is 15. An amount of 80.0 g of Y is dissolved in 190 mL of water. a) What weight of Y would be removed from water with a single extraction with 190-mL of methylene chloride? Show calculations and report to 1 decimal place. b) What weight of Y would be removed from water (the original solution) with two successive extractions with 95-mL portions each of methylene chloride? Show calculations and report to 1 decimal place.
Name: Date: Section: Report Sheet: Determination of Molecular Weight Through Freezing Point Depression Analysis 30 Part A: Volume of Cyclohexane mL Mass of cyclohexane (d =0.7786 g/mL) 6.0 °C Freezing Point of cyclohexane 128 Part B: Molecular Weight of Naphthalene g/mol 0.23 Mass of Sample 1 Moles of Sample 1 Mass of Sample 2 0.18 Moles of Sample 2 Moles m (mol/kg solvent) T,('C) AT, Naphthalene 5.1 TRIAL 1 5.3 TRAIL 2 k, Cyclohexane CC/m) Part C: Unknown Number 0.21 0.19 Mass of Sample 1 Mass of Sample 2 Grams Unknown T,('C) AT 0.21 4.8 TRIAL 1 0.19 5.0 TRAIL 2 Moles UnknownTRIAL 1 (moles) TRIAL 2 (moles) TRIAL 1 (g/mol)TRIAL 2 (g/mol) Molecular Weight Unknown (g/mol) AVERAGE MOLECULAR WEIGHT OF UNKNOWN
Kindly answer no.3 only. Complete answer with solution.
1. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1500 M CaCh2 Solve for the concentration of the stock solution in M. 2. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1200 g CaCl2 Solve for %w/v CaCl2 in the stock solution.
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