need help with #12 Imass of isopropanol (C3H§O).C =12.011 amu x3=36.033 amuH=1,008 annu x 8=8.064 amo= 60.096glmolO= 15,999 amu x 1=15.999 amu9. Calculate the number of moles of isopropanol added to your ice solution. Pureisopropanol is 16.64 M.Molarity (o. of moles x I000)volume of solutionno. of moles :(molaritysvol of sol.)/1000>C16.64 Mx 70 ml000El.165 moles10. Calculate the volume of water in your melted ice and isopropanol solution.Volume after ice melted - volume of Isopropanol70-50=(20ml11. Calculate the mass of water in the solution. (Recall that 1.0 mL of pure water has a massof 1.0 g) Convert your answer to kg.Volume of isopropanol:50 m!Total volume of liquid in tumbler: 7omlVolume of uwater = 70-50 ml=20ml20.0ml-20.0g → 20.og* 1 ha .02 ka12. Recall that the freezing point is depressed by -1.86 °C per mole of solute per kg of water.Calculate the theoretical freezing point of the mixture you created.13. Compare the actual to the theoretical.

Answered: Image /qna-images/answer/4675fee1-84ea-4c20-a032-7685a788d3b7.jpg

Answered: 12. Recall that the freezing point is…

12. Recall that the freezing point is depressed by -1.86 °C per mole of solute per kg of water. Calculate the theoretical freezing point of the mixture you created.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter13: The Chemistry Of Solutes And Solutions

Section: Chapter Questions

Problem 107QRT

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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

need help with #12

Imass of isopropanol (C3H§O).
C =12.011 amu x3=36.033 amu
H=1,008 annu x 8=8.064 amo
= 60.096glmol
O= 15,999 amu x 1=15.999 amu
9. Calculate the number of moles of isopropanol added to your ice solution. Pure
isopropanol is 16.64 M.
Molarity (o. of moles x I000)
volume of solution
no. of moles :(molaritysvol of sol.)/1000>C16.64 Mx 70 ml000
El.165 moles
10. Calculate the volume of water in your melted ice and isopropanol solution.
Volume after ice melted - volume of Isopropanol
70-50=(20ml
11. Calculate the mass of water in the solution. (Recall that 1.0 mL of pure water has a mass
of 1.0 g) Convert your answer to kg.
Volume of isopropanol:50 m!
Total volume of liquid in tumbler: 7oml
Volume of uwater = 70-50 ml=20ml
20.0ml-20.0g → 20.og
* 1 ha .02 ka
12. Recall that the freezing point is depressed by -1.86 °C per mole of solute per kg of water.
Calculate the theoretical freezing point of the mixture you created.
13. Compare the actual to the theoretical.

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