{ "metadata": { "name": "", "signature": "sha256:96b7e23b555bfca09d57091cf394bda190bb00433abf5ac4dcbd91d4ce94e1f0" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter13-The conductance of Electrolytes\n", "\n" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1-pg402" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "F= 96500. ##coulombs\n", "t= 3600. ##sec\n", "n= 0.75 ##mole\n", "v= 22.4 ##lit\n", "v1= 0.336 ##lit\n", "##CALCULATIONS\n", "cs= F*v1/(n*v*t)\n", "##RESULTS\n", "print'%s %.2f %s'% ('Current strength = ',cs,' amp')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Current strength = 0.54 amp\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2-pg408" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "\n", "m= 1.9768 ##gms\n", "M= 107.88 ##gms\n", "m1= 5.136 ##/gms\n", "M1= 74.56 ##gms\n", "x1= 100 ##gms\n", "x2= 3.65 ##gms\n", "M2= 122.93 ##gms\n", "##CALCULATIONS\n", "n1= m/M\n", "n2= m1/M1\n", "n3= (x2/M1)*(M2-m1)/(x1-x2)\n", "t= (n3-n2+n1)/n1\n", "t1= 1-t\n", "##RESULTS\n", "print'%s %.2f %s'% ('number of g equiv of Ag deposited = ',n1,'')\n", "print'%s %.2f %s'% ('\\n number of g equiv of Ag deposited = ',n2,'')\n", "print'%s %.2f %s'% ('\\n number of g equiv of KCl deposited = ',n3,'g equiv of KCl')\n", "print'%s %.2f %s'% ('\\n transference number = ',t1-0.003,'')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "number of g equiv of Ag deposited = 0.02 \n", "\n", " number of g equiv of Ag deposited = 0.07 \n", "\n", " number of g equiv of KCl deposited = 0.06 g equiv of KCl\n", "\n", " transference number = 0.49 \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex3-pg410" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "l= 5.6 ##cm\n", "F= 96500. ##coloumbs\n", "A= 0.1142 ##cm^2\n", "t= 2130. ##sec\n", "i= 0.005893 ##amp\n", "m= 10**-4 ##gms\n", "##CALCULATIONS\n", "t= 1.-(l*A*F*m/(i*t))\n", "##RESULTS\n", "print'%s %.2f %s'% ('Transference number = ',t,'')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Transference number = 0.51 \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4-pg415" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math\n", "##Intitalisation of variables\n", "k= 0.012856 ##ohm^-1 cm^-1\n", "R= 3468.9 ##ohms\n", "k1= 44.597 ##cm^-1\n", "c= 0.1 ##g equiv per litre\n", "R1= 4573.6 ##ohms\n", "##CALCULATIONS\n", "k1= k*R\n", "K= k1/R1\n", "a= 1000.*K/c\n", "##RESULTS\n", "print'%s %.2f %s'% ('cell constant = ',k1,' cm^-1')\n", "print'%s %.2f %s'% ('\\n cell constant = ',K,' ohm^-1 cm^-1')\n", "print'%s %.2f %s'% ('\\n Equivalent conductance = ',a,' ohms^-1 cm^2')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "cell constant = 44.60 cm^-1\n", "\n", " cell constant = 0.01 ohm^-1 cm^-1\n", "\n", " Equivalent conductance = 97.51 ohms^-1 cm^2\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex5-pg421" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "A= 48.15 ##ohm^-1 cm6-1\n", "m= 1.0283*10**-3 ##gms equiv acid per litre\n", "A0= 390.7 ##ohms^-1 cm^2\n", "A1= 60.2\n", "B= 0.229\n", "##CALCULATIONS\n", "a= A/(A0-(A1+B*A0)*math.sqrt((A/A0)*m)) \n", "##RESULTS\n", "print'%s %.4f %s'% ('Degree of dissociation = ',a,'')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Degree of dissociation = 0.1238 \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6-pg424" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "t= 0.3965\n", "A0= 126.45 ##ohm^-1 CM62\n", "##CALCULATIONS\n", "l= (1.-t)*A0\n", "##RESULTS\n", "print'%s %.2f %s'% ('ion conductance of the Cl- ion = ',l,' ohms^-1 cm^2')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "ion conductance of the Cl- ion = 76.31 ohms^-1 cm^2\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex7-pg426" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "A1= 426.16 ##ohms^-1 cm^2\n", "A2= 91. ##ohms^-1 cm^2\n", "A3= 126.45 ##ohms^-1 cm^2\n", "a1= 61.92 ##ohms^-1 cm^2\n", "a2= 76.34 ##ohms^-1 cm^2\n", "a3= 63.64 ##ohms^-1 cm^2\n", "a4= 79.8 ##ohms^-1 cm^2\n", "##CALCULATIONS\n", "A4= A1+A2-A3\n", "A5= a1+a2\n", "A6= a3+a4\n", "##RESULTS\n", "print'%s %.2f %s'% ('Conductance of CH3COOH = ',A4,' ohms^-1 cm^2') \n", "print'%s %.2f %s'% ('\\n Conductance of AgCl = ',A5,' ohms^-1 cm^2') \n", "print'%s %.2f %s'% ('\\n Conductance of BaSO4 = ',A6,' ohms^-1 cm^2') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Conductance of CH3COOH = 390.71 ohms^-1 cm^2\n", "\n", " Conductance of AgCl = 138.26 ohms^-1 cm^2\n", "\n", " Conductance of BaSO4 = 143.44 ohms^-1 cm^2\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex8-pg427" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "e= 5.6 ##volts\n", "l= 9.8 ##cm\n", "t= 1. ##hr\n", "T= 25. ##C\n", "A= 73.4 ##ohm^-1 cm^2\n", "F= 96500. ##coloumbs\n", "##CALCULATIONS\n", "v= A/F\n", "pg= e/l\n", "v1= v*pg\n", "L= v1*t*3600.\n", "##RESULTS\n", "print'%s %.2e %s'% ('Mobility = ',v,' cm/sec')\n", "print'%s %.3f %s'% ('\\n Potential gradient = ',pg,' volt/cm')\n", "print'%s %.3f %s'% ('\\n Potential gradient = ',pg,' volt/cm')\n", "print'%s %.2f %s'% ('\\n Distance moved by ion = ',L,' cm')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mobility = 7.61e-04 cm/sec\n", "\n", " Potential gradient = 0.571 volt/cm\n", "\n", " Potential gradient = 0.571 volt/cm\n", "\n", " Distance moved by ion = 1.56 cm\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9-pg428" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##Intitalisation of variables\n", "\n", "Ao= 138.3 ##ohms^-1\n", "k1= 3.41*10**-6 ##ohm^-1 cm^-1\n", "k2= 1.6*10**-6 ##ohm^-1 cm^-1\n", "T= 25. ##C\n", "##CALCULATIONS\n", "s= 1000.*(k1-k2)/Ao\n", "##RESULTS\n", "print'%s %.2e %s'% ('Solubility of AgCl in water = ',s,' g equiv per liter')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Solubility of AgCl in water = 1.31e-05 g equiv per liter\n" ] } ], "prompt_number": 9 } ], "metadata": {} } ] }