A solution was prepared by adding 3.85g of HNO2(MM =45 g/mol) in 53.2 mL of water. The freezing point of the solution was measured to be -3.5°C. Whacan be said about the solution?kf = 1.86 °C/m; ATf= kf *m*i ; ATf= Tr(solv) - TF(soln)%3D%3DO A. all ions in solutionsOB. all molecules in solutionsO C. mostly ions and a few molecules in solutionO D. mostly molecules and a few ions in solutionO E. equal amounts of molecules and ions in solution

Answered: A solution was prepared by adding 3.85g…

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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A solution was prepared by adding 3.85g of HNO2(MM =45 g/mol) in 53.2 mL of water. The freezing point of the solution was measured to be -3.5°C. Wha
can be said about the solution?
kf = 1.86 °C/m; ATf= kf *m*i ; ATf= Tr(solv) - TF(soln)
%3D
%3D
O A. all ions in solutions
OB. all molecules in solutions
O C. mostly ions and a few molecules in solution
O D. mostly molecules and a few ions in solution
O E. equal amounts of molecules and ions in solution

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19. Given:

Mass of the solute=3.85

Molar mass =45 g/mol

Moles of solute = (3.85 /45 ) mol

Mass of the solvent= 53.2 mL x 1.00 g/mL=53.2 g [Density of water = 1.00 g/mL]

Mass of the solvent=53.2 / 1000=0.0532 kg

Molality, m =(3.85 /45 ) / 0.0532 kg=1.61 m

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