Given: Mass of lauric acid (g): 8.000g Mass of benzoic acid (g): 1.027g Freezing temperature of pure lactic acid(°C): 43.3°C Freezing point of the benzoic acid-lauric acid mixture(°C): 39.8°C m of lauric acid: 1.8m Mols of benzoic acid solute: .0072mol Question: Determine the accepted molecular weight of benzoic acid from its formula, C6H5CO2H.

Given: Mass of lauric acid (g): 8.000g Mass of benzoic acid (g): 1.027g Freezing temperature of pure lactic acid(°C): 43.3°C Freezing point of the benzoic acid-lauric acid mixture(°C): 39.8°C m of lauric acid: 1.8m Mols of benzoic acid solute: .0072mol Question: Determine the accepted molecular weight of benzoic acid from its formula, C6H5CO2H.

EBK A SMALL SCALE APPROACH TO ORGANIC L

4th Edition

ISBN:9781305446021

Author:Lampman

Publisher:Lampman

Chapter13: Isolation Of Eugenol From Clov

Section: Chapter Questions

Problem 9Q

See similar textbooks

Similar questions

A solution is prepared by dissolving 155.0 mL of methanol (CH3OH, density = 0.791 g/mL) in 250 mL of water (density 1.00 g/mL) to give a final solution volume of 350.0 mL. Calculate the (a) molarity of the solution, (b) %wt/wt, (c) %wt/vol, (d) %vol/vol
Calculate the percent composition of sand in the mixture using the mass of the recovered sand Calculate the theoretical mass of salt in the mixtue using the mass of the recovered sand Calculate the theoretial percent composition of salt in the mixture Weight of mixture: 1.841 Weight of salt: 1.319 Weight of sand: 0.376 Weight of sand after heating: 0.376
Consider a blend of mass 5g formed from 2.2 g of C12H24 and 2.8g of C998H1000 paraffin's:, What are Mn and Mw of the blend?
Given Active Ingredient: precipitated sulfur (ointment) Raw Materials: 500 g calcium polysulphide and 1.5 kg hydrochloric acid Actual Yield: 343.4g precipitated sulfur Formulation: 250 mg per jar Dosage form: Ointment packaging:100 jars per box Synthesis and Packaging (Need answer)- Balanced Chemical Equation:- % composition by mass of each compound:- Mass to Mass Stoichiometry Calculation:- Limiting Reagent:- Excess Reagent:- Amount (g) in excess: % Yield:- Number of dosage form and packaging that can be produced from stoichiometric solution:
Mass of empty beaker=16.26Volume of CuSO4 5H2O=30mLMolarity of CuSO4 5H2O used=0.542Moles of CuSO4 5H2O=0.016Mass of aluminum foil used=0.26gMass of beaker and copper product=52.12gMass of copper metal product=35.86gMoles of Al used = 0.009636 molMoles of CuSO4 5H2O used=0.016mol Find the moles of copper product based on moles of Al.Find the moles of copper product based on moles of CuSO4 5H2OFind the limiting reactant in grams
Stock iron(II) solution (200Ug mL-1 Fe) ferrous ammonium sulfate hexahydrate mass= 0.1437g, transfer it to a 100 ml beaker. add 15 ml approx of water and 15m1 'approx of dilute sulphuric acid (2M H2SO.). then transfer FeII to 100 ml flask makeup to the mark with water. calculate the moles of ferrous ammonium sulfate hexahydrate solution in unit ug/mL.
About 200 grams of glucose was fermented with 20 grams of yeast inside a fermenting vessel for 2 weeks. Calculate the percent yield of ethyl alcohol in mL or %(v/v) if the actual yield of the fermentation set-up is at 84.50 mL alcohol. Assume the density of ethanol is at 0.789 g/mL. C6H12O6 + H2O --> 2C2H5OH + 2CO2 Molar Masses: Glucose: 180.156 g/mole; Ethanol: 46.07 g/mole Water: 18.00 g/mole Report your answers in two decimal places. Please include the unit. For example: 55.33%
Dapsone tablets, labelled 50 mg has an average tablet weight of 0.6509 g. 2.0g of powdered tablets reacted (see reaction below) with 26.05 mL of 0.0506N sodium nitrite (NaNO2) solution. Determine the amount of Dapsone as a percentage of the Label Strength (or Label Claim). Report your answer to 1 decimal place. Given: Dapsone molar mass, 248.302 g/mol Dapsone structure Reaction: Sodium nitrite (NaNO2) reacts with hydrochloric acid (HCl) to generate nitrous acid (HNO2). NaNO2 + HCl ® HNO2 + NaCl Nitrous acid (HNO2) then reacts with each amino group on Dapsone, and an indicator indicates the endpoint. R-NH2 (full structure shown above) + HNO2 + HCl ® R-N2Cl + 2H2O
A) An aqueous solution of sodium nitrate, NaO3, is made by dissolving 21.6 grams of sodium nitrate in sufficient water in a 500. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of sodium nitrate in the solution? Weight/volume percentage = _______% B) An aqueous solution of ammonium fluoride, NH4F, is made by dissolving 32.4 grams of ammonium fluoride in sufficient water in a 500. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of ammonium fluoride in the solution? Weight/volume percentage = ________%
Determination of the Amount of water, mass of crucible + dehydrated Sample is 16.2265g, mass of dehydrated Sample is 1.3031g, mass of water is 0.3587g, calculate moles of water.
A solute, having a mass of 4.5 lbₘ and density of 1.5 g/cm³ at room temperature is mixed with a solvent with a mass of 350 oz and density of 71 lbₘ/ft³ at room temperature. Determine the percent volume of solute.
Calculate the percent composition of sand in the mixture using the mass of the recovered sandCalculate the theoretical mass of salt in the mixtue using the mass of the recovered sandCalculate the theoretial percent composition of salt in the mixtureCalculate the percent error for the recovered saltCalculate the total mass of the recovered solids Weight of mixture: 1.841Weight of salt: 1.319Weight of sand: 0.376Weight of sand after heating: 0.376