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(NH),CO). The normal boiling point of a certain liquid X is 132.3 °C, but when 26. g of urea are dissolved in 750. g of X, it is found that the solution boils at 133.5 °C instead. Use this information to calculate the molal boiling point elevation constant K, of X. Round your answer to 2 significant digits. -1 l °C- mol kg planation Check 2022 McGrae Hill LLC AlE Rights Reserved Terms of Use Privacy Center Accessibility tv

(NH),CO). The normal boiling point of a certain liquid X is 132.3 °C, but when 26. g of urea are dissolved in 750. g of X, it is found that the solution boils at 133.5 °C instead. Use this information to calculate the molal boiling point elevation constant K, of X. Round your answer to 2 significant digits. -1 l °C- mol kg planation Check 2022 McGrae Hill LLC AlE Rights Reserved Terms of Use Privacy Center Accessibility tv

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
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(NH),CO). The normal boiling point of a certain liquid X is 132.3 °C, but when 26. g of urea are dissolved in 750. g of X, it is found that the solution boils at 133.5 °C instead. Use this information to calculate the molal boiling point elevation constant K, of X. Round your answer to 2 significant digits. -1 °C-mol kg Eplanation Check 2022 McGrae Hill LLC AlE Rights Reseved Terms of Use Privacy Center Accessibility tv
Transcribed Image Text:(NH),CO). The normal boiling point of a certain liquid X is 132.3 °C, but when 26. g of urea are dissolved in 750. g of X, it is found that the solution boils at 133.5 °C instead. Use this information to calculate the molal boiling point elevation constant K, of X. Round your answer to 2 significant digits. -1 °C-mol kg Eplanation Check 2022 McGrae Hill LLC AlE Rights Reseved Terms of Use Privacy Center Accessibility tv
Expert Solution
Step 1

Given that: 

Mass of urea = 26.g

Molar mass of urea = 60.0 g/mol 

Moles of urea = 26.g÷60.0g/mol = 0.433mol 

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