From 67068710030ddd6b6c809518c34af2e04e0bf7ca Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 16:05:18 +0530 Subject: add book --- Chemistry/Chapter_12.ipynb | 446 +++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 446 insertions(+) create mode 100755 Chemistry/Chapter_12.ipynb (limited to 'Chemistry/Chapter_12.ipynb') diff --git a/Chemistry/Chapter_12.ipynb b/Chemistry/Chapter_12.ipynb new file mode 100755 index 00000000..c565711d --- /dev/null +++ b/Chemistry/Chapter_12.ipynb @@ -0,0 +1,446 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12:Physical Properties of Solutions" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.2,Page no:518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "msolute=0.892 #mass of solute, g\n", + "msolvent=54.6 #mass of solvent, g\n", + "\n", + "#Calculation\n", + "percent=msolute/(msolute+msolvent)*100 #concentration, percent by mass\n", + "\n", + "#Result\n", + "print\"The concentration of KCl solution by mass is :\",round(percent,2),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The concentration of KCl solution by mass is : 1.61 %\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.3,Page no:518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration \n", + "mass=24.4 #mass of H2SO4, g\n", + "M=98.09 #mol maass of H2SO4, g\n", + "\n", + "#Calculation\n", + "n=mass/M #moles of H2SO4\n", + "massH2O=0.198 #mass of H2O, kg\n", + "m=n/massH2O #molality of H2SO4, molal\n", + "\n", + "#Result\n", + "print\"The molality of sulfuric acid solution is :\",round(m,2),\"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The molality of sulfuric acid solution is : 1.26 m\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.4,Page no:520" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#considering 1L solution\n", + "msolution=976 #mass of solution, g\n", + "n=2.45 #moles\n", + "CH3OH=32.04 #mol. mass of CH3OH, g\n", + "\n", + "#Calculation\n", + "msolute=n*CH3OH #mass of solute, g\n", + "msolvent=(msolution-msolute)/1000 #mass of solvent, kg\n", + "m=n/msolvent #molality, molal\n", + "\n", + "#Result\n", + "print\"The molality of CH3OH solution is :\",round(m,2),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The molality of CH3OH solution is : 2.73 m\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.5,Page no:520" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#considering 100g of solution\n", + "percent=35.4 #mass percent of H3PO4\n", + "H3PO4=97.99 #mol mass of H3PO4\n", + "\n", + "#Calculation\n", + "n=percent/H3PO4 #moles of H3PO4\n", + "mH2O=(100-percent)/1000 #mass of solvent\n", + "m=n/mH2O #molality of H3PO4, molal\n", + "\n", + "#Result\n", + "print\"the molality of H3PO4 solution is :\",round(m,2),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the molality of H3PO4 solution is : 5.59 m\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.6,Page no:525" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "c=6.8*10**-4 #solubility of N2 in water, M\n", + "P=1 #pressure, atm\n", + "\n", + "#Calculation\n", + "k=c/P #henry's constant\n", + "#for partial pressure of N2=0.78atm\n", + "P=0.78 #partial pressure of N2, atm\n", + "c=k*P #solubility of N2, M\n", + "\n", + "#Result\n", + "print\"The solubility of N2 gas in water is :%.1e\"%c,\"M\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The solubility of N2 gas in water is :5.3e-04 M\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.7,Page no:527" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "H2O=18.02 #mol mass of H2O, g\n", + "V=460 #volume of water, mL\n", + "glucose=180.2 #mol. mass of glucose, g\n", + "mass=218 #mass of gllucose, g\n", + "\n", + "#Calculation\n", + "n1=V/H2O #moles of water\n", + "n2=mass/glucose #moles of glucose\n", + "x1=n1/(n1+n2) #mole fraction of water\n", + "P=31.82 #vapor pressure of pure water, mmHg\n", + "P1=x1*P #vapor pressure afteraddition of glucose, mmHg\n", + "#Result\n", + "print\"Vapor pressure is:\",round(P1,1),\"mm Hg\"\n", + "print\"The vapor pressure lowering is :\",round(P-P1,1),\"mmHg\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vapor pressure is: 30.4 mm Hg\n", + "The vapor pressure lowering is : 1.4 mmHg\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.8,Page no:532" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "mH2O=2.505 #mass of H2O, kg\n", + "mEG=651 #mass of EG, g\n", + "EG=62.07 #mol mass of EG, g\n", + "\n", + "#Calculation\n", + "n=mEG/EG #moles of EG\n", + "m=n/mH2O #molality of EG\n", + "Kf=1.86 #molal freezing point depression constant, C/m\n", + "deltaTf=Kf*m #depression in freezing point, C\n", + "Kb=0.52 #molal boiling point elevation constant, C/m\n", + "deltaTb=Kb*m #elevation in boiling point, C\n", + "\n", + "#Result\n", + "print\"The depression in freezing point is\",round(deltaTf,2),\"C\"\n", + "print\"Elevation in boiling point is :\",round(deltaTb,1),\"C\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The depression in freezing point is 7.79 C\n", + "Elevation in boiling point is : 2.2 C\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.9,Page no:536" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "pie=30 #osmotic pressure, atm\n", + "R=0.0821 #gas constant, L atm/K mol\n", + "T=298 #temp., K\n", + "\n", + "#Calculation\n", + "M=pie/(R*T) #molar concentration, M\n", + "\n", + "#Result\n", + "print\"The molar concentration is :\",round(M,2),\"M\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The molar concentration is : 1.23 M\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.10,Page no:537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "deltaTf=1.05 #depression in freezing point, C\n", + "Kf=5.12 #molal freezing point depression constant\n", + "\n", + "#Calculation\n", + "m=deltaTf/Kf #molality of solution, molal\n", + "mbenzene=301/1000.0 #mass of benzene, kg\n", + "n=m*mbenzene #moles of sapmle\n", + "msample=7.85 #mass of sample, g\n", + "molarmass=msample/n #molar mass of sample, g/mol\n", + "\n", + "#Result\n", + "print\"The molar mass of the sample is :\",round(molarmass),\"g/mol \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The molar mass of the sample is : 127.0 g/mol \n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.11,Page no:538" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R=0.0821 #gas constant, L atm/K mol\n", + "T=298 #temp, K\n", + "pie=10/760.0 #osmotic pressure, atm\n", + "\n", + "#Calculation\n", + "M=pie/(R*T) #molarity of the solution, M\n", + "#taking 1L of solution\n", + "mass=35 #mass of Hg, g\n", + "n=M #moles\n", + "molarmass=mass/n #molar mass of hemoglobin, g/mol\n", + "\n", + "#Result\n", + "print\"The molar mass of the hemoglobin is :%.2e\"%molarmass,\"g/mol\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The molar mass of the hemoglobin is :6.51e+04 g/mol\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:12.12,Page no:540" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R=0.0821 #gas constant, L atm/K mol\n", + "T=298 #temp, K\n", + "pie=0.465 #osmotic pressure, atm\n", + "M=0.01 #molarity of the solution, M\n", + "\n", + "#Calculation\n", + "i=pie/(M*R*T) #vant hoff factor of KI\n", + "\n", + "#Result\n", + "print\"The vant hoff factor of KI at 25 C is :\",round(i,2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The vant hoff factor of KI at 25 C is : 1.9\n" + ] + } + ], + "prompt_number": 48 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit