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(10 points) In an experiment, a student dissolves 24.6g of LiBr in 0.36kg of solvent. What is the molality of this solution? Show your work. Your Answer: 24.6g of LiBr 0.36kg solvent. molar mass of LiBr = MM(LiBr) 1x 6.941g/mol + 1x 79.90= 86.841g/mol mass of LiBr/Molar mass of LiBr= 24.6/86.841moles-0.2833 moles Moles of solute LiBr/ mass of solevent 0.2833/0.36= 0.78694444 mol/kg What is the molality of this solution? 0.79mol/kg ANSWER: 1.58 m 24.6g LiBrx- 1 mol LiBr 86.8g -x 2mol ions 1 mol LiBr 0.567 mole LiBr ions 0.36kg -=0.567 mol LiBrions €1.575 mLiBr

(10 points) In an experiment, a student dissolves 24.6g of LiBr in 0.36kg of solvent. What is the molality of this solution? Show your work. Your Answer: 24.6g of LiBr 0.36kg solvent. molar mass of LiBr = MM(LiBr) 1x 6.941g/mol + 1x 79.90= 86.841g/mol mass of LiBr/Molar mass of LiBr= 24.6/86.841moles-0.2833 moles Moles of solute LiBr/ mass of solevent 0.2833/0.36= 0.78694444 mol/kg What is the molality of this solution? 0.79mol/kg ANSWER: 1.58 m 24.6g LiBrx- 1 mol LiBr 86.8g -x 2mol ions 1 mol LiBr 0.567 mole LiBr ions 0.36kg -=0.567 mol LiBrions €1.575 mLiBr

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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(10 points) In an experiment, a student dissolves 24.6g of LiBr in 0.36kg of solvent. What is the molality of this solution? Show your work. Your Answer: 24.6g of LiBr 0.36kg solvent. molar mass of LiBr = MM(LiBr) 1x 6.941g/mol + 1x 79.90= 86.841g/mol mass of LiBr/Molar mass of LiBr= 24.6/86.841moles-0.2833 moles Moles of solute LiBr/ mass of solevent 0.2833/0.36= 0.78694444 mol/kg What is the molality of this solution? 0.79mol/kg ANSWER: 1.58 m 24.6g LiBrx- 1 mol LiBr 86.8g -x 2mol ions 1 mol LiBr 0.567 mole LiBr ions 0.36kg -=0.567 mol LiBrions €1.575 mLiBr
Transcribed Image Text:(10 points) In an experiment, a student dissolves 24.6g of LiBr in 0.36kg of solvent. What is the molality of this solution? Show your work. Your Answer: 24.6g of LiBr 0.36kg solvent. molar mass of LiBr = MM(LiBr) 1x 6.941g/mol + 1x 79.90= 86.841g/mol mass of LiBr/Molar mass of LiBr= 24.6/86.841moles-0.2833 moles Moles of solute LiBr/ mass of solevent 0.2833/0.36= 0.78694444 mol/kg What is the molality of this solution? 0.79mol/kg ANSWER: 1.58 m 24.6g LiBrx- 1 mol LiBr 86.8g -x 2mol ions 1 mol LiBr 0.567 mole LiBr ions 0.36kg -=0.567 mol LiBrions €1.575 mLiBr
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