initial commit / add all books

22 files changed, 224 insertions, 0 deletions

diff --git a/3428/CH12/EX6.12.1/Ex6_12_1.sce b/3428/CH12/EX6.12.1/Ex6_12_1.sce
new file mode 100644
index 000000000..dc93bf11c
--- /dev/null
+++ b/3428/CH12/EX6.12.1/Ex6_12_1.sce
@@ -0,0 +1,7 @@
+//Section-6,Example-1,Page no.-P.31
+//To determine the weight of glucose required to prepare 2.0 M.
+clc;
+M=2
+V=1
+x=180*M
+disp(x,'Weight of glucose required to prepare 2.0 M(gm)')
diff --git a/3428/CH12/EX6.12.10/Ex6_12_10.sce b/3428/CH12/EX6.12.10/Ex6_12_10.sce
new file mode 100644
index 000000000..814160cfe
--- /dev/null
+++ b/3428/CH12/EX6.12.10/Ex6_12_10.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-2,Page no.-P.42
+//To calculate the minimum partial pressure of methane to achieve the following condition.
+clc;
+n_M=(21*10^-3)/16
+n_B=100/78
+x_M=n_M/(n_M+n_B)
+K_M=4.27*10^5
+p_M=x_M*K_M
+disp(p_M,'Required partial pressure(Torr)')
diff --git a/3428/CH12/EX6.12.11/Ex6_12_11.sce b/3428/CH12/EX6.12.11/Ex6_12_11.sce
new file mode 100644
index 000000000..b60239971
--- /dev/null
+++ b/3428/CH12/EX6.12.11/Ex6_12_11.sce
@@ -0,0 +1,17 @@
+//Section-6,Example-3,Page no.-P.42
+clc;
+K_B=1.25*10^6
+n_A=1000/18
+n_B=n_A/K_B
+p_B1=0.1*760
+p_B2=5*760
+p_B3=2*760
+n_B1=(n_B*p_B1)
+m1_CO2=n_B1
+disp(m1_CO2,'Molality of CO2(mol/kg)')
+n_B2=(n_B*p_B2)
+m2_CO2=n_B2
+disp(m2_CO2,'Molar conc. of CO2(mol/L)')
+n_B3=(n_B*p_B3)
+m3_CO2=n_B3
+disp(m3_CO2,'Molar conc. of CO2(mol/L)')
diff --git a/3428/CH12/EX6.12.12/Ex6_12_12.sce b/3428/CH12/EX6.12.12/Ex6_12_12.sce
new file mode 100644
index 000000000..94349ff6f
--- /dev/null
+++ b/3428/CH12/EX6.12.12/Ex6_12_12.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-1,Page no.-P.43
+//To find the lowering of freezing point of the solution.
+clc;
+K_f=1.86
+M_m=(12*12)+(22*1)+(11*16)       //Molar mass of sucrose(C_12H_22O_11)(gm/mol)
+n_S=3/M_m
+m_S=n_S/0.1
+dl_Tf=K_f*m_S
+disp(dl_Tf,'Lowering of freezing point of the solution(K)')
diff --git a/3428/CH12/EX6.12.13/Ex6_12_13.sce b/3428/CH12/EX6.12.13/Ex6_12_13.sce
new file mode 100644
index 000000000..a57ef3cd9
--- /dev/null
+++ b/3428/CH12/EX6.12.13/Ex6_12_13.sce
@@ -0,0 +1,11 @@
+//Section-6,Example-5,Page no.-P.48
+//To calculate the quantity of ethyl alcohol required.
+clc;
+dl_Tf=10                  //(K)
+K_f=1.86                  //(Kkgmol^-1)
+M_w=62
+d=1                    //density (assumption)
+V=10                   //Volume(L)
+M=V*d
+W=((dl_Tf*M*M_w)/K_f)*10^-3
+disp(W,'Quantity of ethyl alcohol required(kg)')
diff --git a/3428/CH12/EX6.12.14/Ex6_12_14.sce b/3428/CH12/EX6.12.14/Ex6_12_14.sce
new file mode 100644
index 000000000..90a79d799
--- /dev/null
+++ b/3428/CH12/EX6.12.14/Ex6_12_14.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-6,Page no.-P.49
+//To calculate the molar mass of the compound.
+clc;
+w_B=28
+w_A=750*10^-3
+dl_Tf=5.4
+K_f=30
+m_B=(K_f*w_B)/(dl_Tf*w_A)
+disp(m_B,'Molar mass of the given compound(gmmol^-1)')
diff --git a/3428/CH12/EX6.12.15/Ex6_12_15.sce b/3428/CH12/EX6.12.15/Ex6_12_15.sce
new file mode 100644
index 000000000..0013ce673
--- /dev/null
+++ b/3428/CH12/EX6.12.15/Ex6_12_15.sce
@@ -0,0 +1,11 @@
+//Section-6,Example-7,Page no.-P.49
+//To calculate the freezing point of the given aqueous solution.
+clc;
+T_bbar=373
+T_fbar=273
+T_b=373.1
+dl_H=1            //(let) dl_H=(dl_Hvap./dl_Hfus.)
+dl_Tb=(T_b-T_bbar)
+dl_Tf=dl_Tb*((T_fbar)^2/(T_bbar)^2)*dl_H
+disp(dl_Tf,' Freezing point of the given aqueous solution')
+//dl_Tf=dl_Tf*(dl_Hvap./dl_Hfus.)
diff --git a/3428/CH12/EX6.12.16/Ex6_12_16.sce b/3428/CH12/EX6.12.16/Ex6_12_16.sce
new file mode 100644
index 000000000..acf03456d
--- /dev/null
+++ b/3428/CH12/EX6.12.16/Ex6_12_16.sce
@@ -0,0 +1,13 @@
+//Section-6,Example-1,Page no.-P.53
+//To find the molality of the sucrose solution.
+clc;
+R=0.082
+M=342
+V=1
+T=298
+pi=4.82                //Osmotic pressure(atm)
+w=(pi*M*V)/(R*T)         //weight of sucrose(gm)
+w_W=(1015-w)*10^-3        //Weight of water in 1015 gm of sucrose solution(kg)
+n_S=w/M                   //moles of sucrose
+m_S=n_S/w_W
+disp(m_S,'Molality of the sucrose solution(m)')
diff --git a/3428/CH12/EX6.12.17/Ex6_12_17.sce b/3428/CH12/EX6.12.17/Ex6_12_17.sce
new file mode 100644
index 000000000..cd5b1c7ee
--- /dev/null
+++ b/3428/CH12/EX6.12.17/Ex6_12_17.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-2,Page no.-P.53
+//To determine the extent of dilution of the solution.
+clc;
+T_1=283
+T_2=298
+P_1=500
+P_2=105.3
+a=(T_2/T_1)*(P_1/P_2)
+disp(a,'Extent of dilution of the solution')
diff --git a/3428/CH12/EX6.12.18/Ex6_12_18.sce b/3428/CH12/EX6.12.18/Ex6_12_18.sce
new file mode 100644
index 000000000..5e5281f63
--- /dev/null
+++ b/3428/CH12/EX6.12.18/Ex6_12_18.sce
@@ -0,0 +1,8 @@
+//Section-6,Example-3,Page no.-P.54
+//To calculate the molecular mass of Urea.
+clc;
+w_1=10/100
+w_2=1.754/100
+M_1=342
+M_2=(w_2*M_1)/w_1
+disp(M_2,'Molecular mass of Urea(g/mol)')
diff --git a/3428/CH12/EX6.12.19/Ex6_12_19.sce b/3428/CH12/EX6.12.19/Ex6_12_19.sce
new file mode 100644
index 000000000..9d9f12eed
--- /dev/null
+++ b/3428/CH12/EX6.12.19/Ex6_12_19.sce
@@ -0,0 +1,8 @@
+//Section-6,Example-4,Page no.-P.54
+//To calculate the osmotic pressure under the given conditions.
+clc;
+R=0.0821
+B_f=3/(300*R*2)
+T=400
+pi_f=B_f*R*T
+disp(pi_f,'Required osmotic pressure (atm)')
diff --git a/3428/CH12/EX6.12.2/Ex6_12_2.sce b/3428/CH12/EX6.12.2/Ex6_12_2.sce
new file mode 100644
index 000000000..2c34d72f0
--- /dev/null
+++ b/3428/CH12/EX6.12.2/Ex6_12_2.sce
@@ -0,0 +1,8 @@
+//Section-6,Example-2,Page no.-P.31
+//To determine the mass of glycine required to prepare 100 ml of its solution.
+clc;
+M_m=(14*1)+(1*5)+(12*2)+(16*2)
+V=1
+M=0.01
+x=M*100*1000*M_m
+disp(x,'Required  mass of glycine(gm)')
diff --git a/3428/CH12/EX6.12.20/Ex6_12_20.sce b/3428/CH12/EX6.12.20/Ex6_12_20.sce
new file mode 100644
index 000000000..45ea9dd10
--- /dev/null
+++ b/3428/CH12/EX6.12.20/Ex6_12_20.sce
@@ -0,0 +1,11 @@
+//Section-6,Example-5,Page no.-P.55
+//To calculate the osmotic pressure under the given conditions.
+clc;
+T=27+273
+B_Sucrose=(0.17/342)/(50/100)
+B_Glucose=(0.18/180)/(50/1000)
+B_Urea=(0.06/60)/(50/1000)
+B=B_Sucrose+B_Glucose+B_Urea
+R=0.0821
+pi=B*R*T
+disp(pi,'Required osmotic pressure(atm)')
diff --git a/3428/CH12/EX6.12.21/Ex6_12_21.sce b/3428/CH12/EX6.12.21/Ex6_12_21.sce
new file mode 100644
index 000000000..e8f733488
--- /dev/null
+++ b/3428/CH12/EX6.12.21/Ex6_12_21.sce
@@ -0,0 +1,8 @@
+//Section-6,Example-6,Page no.-P.55
+//To calculate the molecular weight of the polymer.
+clc;
+pi=(2.4*0.88)/(13.6*760)               //Osmotic pressure(atm)
+R=0.0821                              //LatmK^-1mol^-1
+T=273+37
+M=(2.5/pi)*R*T
+disp(M,'Molecular weight of the given polymer')
diff --git a/3428/CH12/EX6.12.22/Ex6_12_22.sce b/3428/CH12/EX6.12.22/Ex6_12_22.sce
new file mode 100644
index 000000000..46ccc1a9a
--- /dev/null
+++ b/3428/CH12/EX6.12.22/Ex6_12_22.sce
@@ -0,0 +1,10 @@
+//Section-6,Example-7,Page no.-P.55
+//To find the concentration of glucose solution.
+clc;
+pi_Blood=7.65
+R=0.0821
+T=310
+B_Glucose=pi_Blood/(R*T)
+M_w=180        //Molecular weight of Glucose(g/mol)
+C=B_Glucose*M_w
+disp(C,'Concentration of glucose solution(gm/L)')
diff --git a/3428/CH12/EX6.12.3/Ex6_12_3.sce b/3428/CH12/EX6.12.3/Ex6_12_3.sce
new file mode 100644
index 000000000..ad40293a0
--- /dev/null
+++ b/3428/CH12/EX6.12.3/Ex6_12_3.sce
@@ -0,0 +1,7 @@
+//Section-6,Example-3,Page no.-P.32
+//To find the Mole fraction of glycine in given aqueous solution.
+clc;
+m_g=0.1*1        //moles of glycine
+m_w=10^3/18         //moles of water in 1 kg of water
+m_f=m_g/(m_g+m_w)
+disp(m_f,'Mole fraction of glycine')
diff --git a/3428/CH12/EX6.12.4/Ex6_12_4.sce b/3428/CH12/EX6.12.4/Ex6_12_4.sce
new file mode 100644
index 000000000..0eb3604cc
--- /dev/null
+++ b/3428/CH12/EX6.12.4/Ex6_12_4.sce
@@ -0,0 +1,24 @@
+//Section-6,Example-4,Page no.-P.33
+clc;
+N_2=6.61
+V_2=1000
+V_1=180
+N_1=(N_2*V_2)/V_1
+disp(N_1,'Normality of conc.H_2SO_4(N)')
+N_2=6.61           //Normality of dil.H_2SO_4
+M_1=N_2/2
+disp(M_1,'Molarity of dil.H_2SO_4(M)')
+W_H=6.61*49       //Weight of H_2SO_4 actually contained in the solution(gm)
+W_CH=180*1.84     //Weight of 180 ml of conc. H_2SO_4(gm)
+pr_H=(W_H/W_CH)*100
+disp(pr_H,'% of H_2SO_4 by weight in conc. H_2SO_4')
+W_H1=49*6.61          //Weight of H_2SO_4 on 1L of dil.solution
+y=W_H1/(1000*1.198*0.01)
+disp(y,'% of H_2SO_4 by weight in diluted H_2SO_4')
+W_H2O=1198-323.85         //Weight of water(gm)
+m=M_1/(W_H2O*10^-3)
+disp(m,'Molality of diluted solution(m)')
+M_H2SO4=W_H1/98
+M_Water=W_H2O/18
+M_F=M_H2SO4/(M_H2SO4+M_Water)
+disp(M_F,'Mole fraction of H_2SO_4 in diluted solution')
diff --git a/3428/CH12/EX6.12.5/Ex6_12_5.sce b/3428/CH12/EX6.12.5/Ex6_12_5.sce
new file mode 100644
index 000000000..bd1c6e30d
--- /dev/null
+++ b/3428/CH12/EX6.12.5/Ex6_12_5.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-1,Page no.-P.35
+//To calculate the total volume of mixture of 50gm of ethylalcohol and 50gm of water at 25degree Celcius.
+clc;
+n_1=50/46           //Moles of C_2H_5OH
+n_2=50/18           //Moles of H_2O
+V_1=55              //Volume of C_2H_5OH
+V_2=18              //Volume of H_2O
+V=(n_1*V_1)+(n_2*V_2)
+disp(V,'Total volume of mixture(ml)')
diff --git a/3428/CH12/EX6.12.6/Ex6_12_6.sce b/3428/CH12/EX6.12.6/Ex6_12_6.sce
new file mode 100644
index 000000000..ed5d5c438
--- /dev/null
+++ b/3428/CH12/EX6.12.6/Ex6_12_6.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-1,Page no.-P.38
+//To find the change in chemical potential of the substance in the given condition.
+clc;
+R=8.314
+T=298
+P=0.5
+P_0=1
+mu_1=R*T*log(P/P_0)         //mu_1=mu-mu_0
+disp(mu_1,'Change in chemical potential of the substance(J/mol)')
diff --git a/3428/CH12/EX6.12.7/Ex6_12_7.sce b/3428/CH12/EX6.12.7/Ex6_12_7.sce
new file mode 100644
index 000000000..7a6edbc4e
--- /dev/null
+++ b/3428/CH12/EX6.12.7/Ex6_12_7.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-1,Page no.-P.38
+//To find the partial vapour pressure of benzene in the solution.
+clc;
+n_B=1000/78
+n_n=1.5
+p_Bbar=94.6
+x_B=n_B/(n_B+n_n)
+p_B=x_B*p_Bbar
+disp(p_B,'Partial vapour pressure of benzene(Torr)')
diff --git a/3428/CH12/EX6.12.8/Ex6_12_8.sce b/3428/CH12/EX6.12.8/Ex6_12_8.sce
new file mode 100644
index 000000000..ce92b4498
--- /dev/null
+++ b/3428/CH12/EX6.12.8/Ex6_12_8.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-1,Page no.-P.40
+//To find by how much is the chemical potential of benzene reduced for the given conditions.
+clc;
+x_B=0.10
+x_A=1.0-0.10
+R=8.314
+T=298
+mu=R*T*log(x_A)       //mu=mu_A-mu_Abar
+disp(mu,'Required chemical potential(Jmol^-1)')
diff --git a/3428/CH12/EX6.12.9/Ex6_12_9.sce b/3428/CH12/EX6.12.9/Ex6_12_9.sce
new file mode 100644
index 000000000..b96dc37fa
--- /dev/null
+++ b/3428/CH12/EX6.12.9/Ex6_12_9.sce
@@ -0,0 +1,9 @@
+//Section-6,Example-1,Page no.-P.41
+//To calculate the minimum partial pressure of oxygen in the atmosphere that can achieve the given concentration.
+clc;
+n_O2=(4*10^-3)/(32)
+n_H2O=(1.0*10^3)/(18)
+x_O2=n_O2/(n_O2+n_H2O)
+K_O2=3.3*10^7
+p_O2=x_O2*K_O2
+disp(p_O2,'Required partial pressure(Torr)')