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How many grams of Mg(NO3)2 (MM = 148.33 g/mol) would it take to lower the freezing point of (4.64x10^2) grams of water by 4.00°C? The freezing point depression constant of water is Kf = 1.86°C/m. %3D %3D Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.

How many grams of Mg(NO3)2 (MM = 148.33 g/mol) would it take to lower the freezing point of (4.64x10^2) grams of water by 4.00°C? The freezing point depression constant of water is Kf = 1.86°C/m. %3D %3D Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter10: Solutions
Section: Chapter Questions
Problem 40QAP: How many grams of the following nonelectrolytes would have to be mixed with 100.0 g of...
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How many grams of Mg(NO3)2 (MM = 148.33 g/mol) would it take to lower the freezing point of (4.64x10^2) grams of water by 4.00°C? The freezing point depression constant of water is Kf = 1.86°C/m. Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.
Transcribed Image Text:How many grams of Mg(NO3)2 (MM = 148.33 g/mol) would it take to lower the freezing point of (4.64x10^2) grams of water by 4.00°C? The freezing point depression constant of water is Kf = 1.86°C/m. Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.
When (8.09x10^1) g of a non-electrolyte is dissolved in (5.9500x10^2) g of a solvent (with Kp = 0.416°C/m) the boiling point of the solution is 1.50°C higher than the boiling point of the pure solvent. What is the molar mass (in g/mol) of the non- electrolyte solute? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.
Transcribed Image Text:When (8.09x10^1) g of a non-electrolyte is dissolved in (5.9500x10^2) g of a solvent (with Kp = 0.416°C/m) the boiling point of the solution is 1.50°C higher than the boiling point of the pure solvent. What is the molar mass (in g/mol) of the non- electrolyte solute? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible.
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