removing problem statements

1 files changed, 0 insertions, 23 deletions

diff --git a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb
index 48ad85eb..23e275b1 100644
--- a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb
+++ b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb
@@ -27,38 +27,26 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "# Calculate Mass Fraction, Mole fraction, Molality and PPM.\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "m_i = 10.            #[g]\n",
       "m_w = 990.           #[g]\n",
       "M_i = 342.3          #[g]\n",
       "M_w = 18.            #[g]\n",
       "\n",
-      "# Calculations\n",
-      "# The mass fraction is \n",
-      "# ( mass fraction of sucrose ) = x_i (by mass) = m_i/(sum of all subsmath.tances)\n",
       "\n",
       "x_i = m_i/(m_i+m_w)\n",
       "x_i = x_i*100.      # [in percentage]\n",
       "\n",
-      "# This is also the weight fraction.\n",
-      "# The mole fraction is\n",
-      "# ( mole fraction of sucrose ) = x_j (by mole) = n_i/(sum of number moles of all the subsmath.tances)\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",
-      "# The molality, a concentration unit is widely used in equilibrium calculations, is defined as \n",
-      "# m (molality) = (moles of solute)/(kg of solvent)\n",
       "m = n_i/m_w*1000            #[molal]\n",
-      "# For solutions of solids and liquids (but not gases) ppm almost always means ppm by mass, so \n",
       "x_ppm = x_i*10**(6)/100.     #[ppm]\n",
       "\n",
-      "# Results\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",
@@ -92,36 +80,25 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "# Calculate Mass concentration , Mole concentration and Molarity\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\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",
-      "# Calculations\n",
-      "# In the previous example i.e. example 1.1 the mass was chosen to be 1.00 kg, so that\n",
       "m = 1.00                      #[kg]\n",
       "V = m/d*1000                  #[L]\n",
       "\n",
-      "# The mass concentration is\n",
-      "# m_1 ( mass concentration of sucrose ) = (mass of sucrose)/(volume of solution)\n",
       "m_1 = m_i/V                   #[g/L]\n",
       "\n",
-      "# The mole concentration is \n",
-      "# m_2 ( mole concentration of sucrose ) = (moles of sucrose)/(volume of solution)\n",
       "\n",
       "m_2 = (m_i/M_i)/V             #[mol/L]\n",
       "\n",
-      "# Results\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",
-      "# By the definition of the molarity, molarity is mole concentration of the solute\n",
-      "# so molarity \n",
       "print \" Molarity of the solution is           %f mol/L\"%(m_2)\n"
      ],
      "language": "python",