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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 1 :  Introduction to Equilibrium"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.1  Page: 9"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "m_i = 10.            #[g]\n",
      "m_w = 990.           #[g]\n",
      "M_i = 342.3          #[g]\n",
      "M_w = 18.            #[g]\n",
      "\n",
      "\n",
      "x_i = m_i/(m_i+m_w)\n",
      "x_i = x_i*100.      # [in percentage]\n",
      "\n",
      "n_i = m_i/M_i       # number of moles of sucrose\n",
      "n_w = m_w/M_w       # number of moles of water\n",
      "x_j = n_i/(n_i+n_w)\n",
      "x_j = x_j*100       # [in percentage]\n",
      "\n",
      "m = n_i/m_w*1000            #[molal]\n",
      "x_ppm = x_i*10**(6)/100.     #[ppm]\n",
      "\n",
      "print \" sucrose concentration in terms of the mass fraction is %f%%\"%(x_i)\n",
      "print \" sucrose concentration in terms of the mole fraction is %f%%\"%(x_j)\n",
      "print \" sucrose concentration in terms of the molality is      %f molal\"%(m)\n",
      "print \" sucrose concentration in terms of the ppm is           %f ppm\"%(x_ppm)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " sucrose concentration in terms of the mass fraction is 1.000000%\n",
        " sucrose concentration in terms of the mole fraction is 0.053088%\n",
        " sucrose concentration in terms of the molality is      0.029509 molal\n",
        " sucrose concentration in terms of the ppm is           10000.000000 ppm\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.2  Page: 9"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "T = 20.                       #[C]\n",
      "d = 1.038143/1000*10.**(6)    #[kg/m**(3)]\n",
      "m_i = 10.                     #[g] mass of sucrose\n",
      "M_i = 342.3                   #[g/mol] molecular weight of sucrose\n",
      "\n",
      "m = 1.00                      #[kg]\n",
      "V = m/d*1000                  #[L]\n",
      "\n",
      "m_1 = m_i/V                   #[g/L]\n",
      "\n",
      "\n",
      "m_2 = (m_i/M_i)/V             #[mol/L]\n",
      "\n",
      "print \" Mass concentration of the solution is %f g/L\"%(m_1)\n",
      "print \" Mole concentration of the solution is %f mol/L\"%(m_2)\n",
      "\n",
      "print \" Molarity of the solution is           %f mol/L\"%(m_2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Mass concentration of the solution is 10.381430 g/L\n",
        " Mole concentration of the solution is 0.030328 mol/L\n",
        " Molarity of the solution is           0.030328 mol/L\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
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