In [1]:

```
# solution
# Variables
# NH4Cl
M = 14+4+35.5 # [g] (molar mass of NH4Cl)
n=5 # [mol]
# Calculation
m = M*n # [g]
# Result
print "5 mol of NH4Cl = ",m," [g]"
```

In [2]:

```
# solution
# Variables
# CuSO4.5H2O
M1 = 159.5 # [g] (molar mass of CuSO4)
M2 = 159.5+5.*(2.+16) # (molar mass of CuSO4.5H2O)
m = 499.
# Calculation
n = m/M2 #[mol]
# Result
print "In the formula CuSO4.5H2O, the moles of CuSO4 is one hence,\
the equivalent moles of CuSO4 in the crystal is ",n,"."
```

In [3]:

```
# solution
# Variables
# K2CO3
m = 117. # [kg] (wt of K)
Mk = 39. # [g] (at wt of K)
# Calculation
a = m/Mk # [kg atoms]
# 1 mol of K2CO3 contains 2 atoms of K
n = a/2. # [kmol] (moles of K2CO3)
# Result
print " ",n," kmol of K2CO3 contains 117 kg of K."
```

In [4]:

```
# solution
# Variables
# BaCl2
M = 137.3+2*35.5 # [g] (molar mass of BaCl2)
m = 416.6 # [g]
# Calculation
n = m/M # [mol]
N = n*6.022*10**23 # (no. of atoms)
# Result
print "Atoms present in 416.6 g BaCl2 = ",N,""
```

In [5]:

```
# solution
# Variables
print "a ",
#PO4 radical
M = 31+4*16. #[g]
V = 3. # (valence of PO4)
# Calculation and Result
eqm = M/V
print "eq. mass of PO4 is ",eqm," [g] "
print "b ",
#Na3PO4
M = 3*23+95. #[g]
V = 3.
eqm = M/V
print "eq. mass of Na3PO4 is ",eqm," [g] "
```

In [6]:

```
# solution
# Variables
# AlCl3
v = 3. # valency of Al ion
# Calculation
eq = 3.*3 # [mol]
# Result
print "no. of equivalents in 3 kmol of AlCl3 is ",eq," keq."
```

In [1]:

```
# solution
# Variables
# (a)
print "a ",
# mass %
m1 = 600. #[kg] (NaCl)
m2 = 200. #[kg] (KCl)
# Calculation
m = m1+m2 # total mass
Wa = (m1/m)*100.
Wb = (m2/m)*100.
# Result
print "mass percentage of NaCl is ",Wa," mass percentage of KCl is ",Wb," "
# (b)
print "b ",
#mol %
M1 = 23+35.5 # molar mass of NaCl
n1 = m1/M1 # no. of moles of NaCl
M2 = 39+35.5 # molar mass of KCl
n2 = m2/M2 # no. of moles of KCl
n = n1+n2
N1 = (n1/n)*100.
N2 = (n2/n)*100.
print "mol percentage of NaCl is ",N1," mol percentage of KCl is ",N2," "
```

In [8]:

```
# solution
# Variables
# CH.35O.35S.14
# mass %
C = 12.0107 #[kg]
H = 1.00794*.35 # [kg]
O = 15.9994*.35 # [kg]
S = 32.065*.14 #[kg]
# Calculation
m = C+H+O+S
m1 = (C/m)*100.
m2 = (H/m)*100.
m3 = (O/m)*100.
m4 = (S/m)*100.
# Result
print "mass percentage of C is ",m1
print " mass percentage of H is ",m2
print " mass percentage of O is ",m3
print " mass percentage of S is ",m4
M = m/(1+.35+.35+.14)
print "molar mass = ",M," kg/kmol."
```

In [9]:

```
# solution
# Variables
# basis 100kg urea
m1 = 45. #[kg] (mass of N present)
Mu = 60. # (molar mass of urea)
m2 = 14*2. #[kg] (mass of N in 1 kmol of urea)
# Calculation
m = (Mu/m2)*m1
# Result
print "The sample contains ",m," percent urea."
```

In [10]:

```
# solution
# Variables
# NaOH
Impurity = 60. # [ppm] SiO2
# Calculation
m = (60/1000000.)*100.
# Result
print "Mass percent of SiO2 is ",m,"."
```

In [11]:

```
# solution
# Variables
Ca = 40.078 # at. wt of Ca
F = 18.9984032 # at wt of F
# Calculation
M1 = 3*Ca +2*(30.97762+(4*15.9994)) # molar mass of Ca3PO4
M2 = Ca +12.0107+3*15.9994 # molar mass of CaCO3
M3 = Ca+2*F # molar mass of CaF2
m1 = 800. #[mg] Ca3PO4
m2 = 200. #[mg] CaCO3
m3 = 5. #[mg] CaF2
n1 = ((3*Ca)/M1)*m1+(Ca/M2)*m2+(Ca/M3)*m3 # [mg] total Ca ions
n2 = (F/M3)*2*5 #[mg] total F ions
# Result
print "Total no. of Ca+ ions is ",n1," and total no. of F- ions is ",n2,"."
```

In [12]:

```
# solution
# Variables
# (a)
print "a ",
# mass %
m1 = 100. #[kg] methanol (basis)
m2 = 64. #[kg] salicylic acid
# Calculation
m = m1+m2 # [kg] mass of solution
w1 = m2/m*100
w2 = 100-w1
# Result
print "mass percent of salicylic acid is ",w1," and mass percent of methanol is ",w2,". "
#(b)
print "b ",
#mole %
M1 = 32. # molar mass of methanol
M2 = 138. #molar mass of salicylic acid
n1 = m1/M1 #[kmol] methanol
n2 = m2/M2 #[kmol] salicylic acid
n = n1+n2
N1 = n1/n*100.
N2 = n2/n*100.
print "Mole percent of methanol is ",N1," and Mole percent of salicylic acid is ",N2,"."
```

In [13]:

```
# solution
# Variables
#mass %
m1 = 13.70 # HCl
m2 = 8.67 # NaCl
m3 = 100. # H2O
# Calculation
m = m1+m2+m3 # mass of solution
w1 = m1/m*100.
w2 = m2/m*100.
w3 = m3/m*100.
# Result
print "mass percent of HCl is ",w1
print "mass percent of NaCl is ",w2
print "and mass percent of H2O is ",w3,". "
M1=36.4609 #HCl
M2=58.4428 #NaCl
M3=18.0153 #H2O
n1=m1/M1 #HCl
n2=m2/M2 #NaCl
n3=m3/M3 #H2O
n=n1+n2+n3
N1=n1/n*100.
N2=n2/n*100.
N3=n3/n*100.
print "Mole percent of HCl is ",N1
print "Mole percent of NaCl is ",N2
print "and Mole percent of H2O is ",N3
```

In [14]:

```
# solution
# Variables
m = 100. #[kg] Lye (basis)
m1 = 73. #[kg] NaOH
M1 = 40. # NaOH
M2 = 62. # Na2O
# Calculation
p = (M2*m1)/(2*M1)
# Result
print "percentage of Na2O in the solution is ",p,"."
```

In [15]:

```
# solution
# Variables
#(CH2OH)3
M = 92. # molar mass of glycerin
C = 600. #[mg/l] glycerin conc.
# Calculation
TOC = (3.*12./92.)*600. #[mg/l]
# by combustion reaction we see 3.5 O2 is required for 1 mol of (CH2OH)3
ThOD = (3.5*32.*600)/92 #[mg/l]
# Result
print "TOC = ",TOC," mg/l"
print "ThOD = ",ThOD," mg/l"
```

In [16]:

```
# solution
# Variables
M1 = 100. # CaCO3
v1 = 2. # valence of CaCO3
# Calculation
eqm1 = M1/v1 # equivalent mass of CaCO3
M2 = 162. # Ca(HCO3)2
v2 = 2.
eqm2 = M2/v2
m = 500. # [mg/l] CaCO3
C1 = (eqm2/eqm1)*m*.6 # [mg/l] conc. of Ca(HCO3)2
M3 = 146.3 # Mg(HCO3)2
v3 = 2.
eqm3 = M3/v3
C2 = (eqm3/eqm1)*m*.4 #[mg/l] conc. of Mg(HCO3)2
# Result
print "Actual concentration of CaHCO32 in the sample\
water is ",C1," mg/l and of MgHCO32 is ",C2," mg/l."
```

In [17]:

```
# solution
# Variables
S = .68 # sulphur content by mass
d = .85 # kg/l
# Calculation
s = (S*d*10**6)/100. # [mg/l] or [ppm]
# Result
print "Sulphur content in LDO is ",s," ppm."
```

In [18]:

```
# solution
# Variables
m1 = 100. #[kg] solution (basis)
m2 = 20. #[kg] NaCl
d = 1.127 #[kg/l]
# Calculation
V = m1/d # volume of 100 kg sol.
n = (m2/58.5)*100. # [mol] NaCl
M = n/V #[M]
v = 1. # valence of NaCl so,
N = M
m = n/(m1-m2) #[mol/kg]
# Result
print "Molarity = ",M,"M"
print "Normality = ",N,"N"
print "Molality = ",m," mol/kg."
```

In [19]:

```
# solution
# Variables
m1 = 100. #[kg] TEA solution (basis)
m2 = 50. #[kg] TEA
M1 = 149. # molar mass of TEA
d = 1.05 #[kg/l]
# Calculation
V = m1/d # volume of 100 kg sol.
n = (m2/M1)*100. # [mol] NaCl
M = n/V #[M]
# Result
print "Molarity of solution = ",M," M."
```

In [20]:

```
# solution
# Variables
m1 = 100. #[kg] MEA solution (basis)
m2 = 20. #[kg] MEA
M1 = 61. # molar mass of MEA
n1 = m2/M1 # [kmol]
C = .206
# Calculation
n2 = C*n1 #[kmol] dissolved CO2
m3 = n2*44 # [kg] mass of CO2
n3 = (m1-m2-m3)/18 #[kmol] water
n = (n2/(n1+n2+n3))*100.
m = (m3/100)*100.
# Result
print "Mass percent of CO2 = ",m," and Mol percent = ",n,"."
```

In [21]:

```
import math
# solution
# Variables
#HOCl
Ma = .1 #molarity
# Calculation
Ka = 9.6*10**-7
C = (Ma*Ka)**.5 # conc. of H+ ions
pH = -math.log10(C)
# Result
print "pH of the sol is ",pH,"."
```

In [2]:

```
# solution
# Variables
n = 100. # [mol] air (basis)
n1 = 21. #[mol] O2
n2 = 78. #[mol] N2
n3 = 1. #[mol] Ar
M1 = 31.9988 # O2
M2 = 28.0134 # N2
M3 = 39.948 # Ar
# Calculation
m1 = n1*M1
m2 = n2*M2
m3 = n3*M3
Ma = (m1+m2+m3)/n
# Result
print "average molar mass of air is ",Ma," g."
```

In [23]:

```
# solution
# Variables
# (a)
print "a ",
n = 100. # [kmol] cracked gas (basis)
n1 = 45. # methane
n2 = 10. # ethane
n3 = 25. # ethylene
n4 = 7. # propane
n5 = 8. # propylene
n6 = 5. # n-butane
M1 = 16.
M2 = 30.
M3 = 28.
M4 = 44.
M5 = 42.
M6 = 58.
# Calculation
m1 = n1*M1
m2 = n2*M2
m3 = n3*M3
m4 = n4*M4
m5 = n5*M5
m6 = n6*M6
m = m1+m2+m3+m4+m5+m6
M = m/n
# Result
print "Average molar mass of gas is ",M," g."
#(b)
print "b ",
# composition
p1 = (m1/m)*100.
p2 = m2*100./m
p3 = m3*100./m
p4 = m4*100./m
p5 = m5*100./m
p6 = m6*100./m
print " GAS Mass Percent "
print " Methane ",p1
print " Ethane ",p2
print " Ethylene ",p3
print " Propane ",p4
print " Propylene ",p5
print " n-Butane ",p6
# (c)
print "c ",
# specific gravity
g = M/28.97
print "Specific gravity is ",g,"."
```

In [24]:

```
# solution
# Variables
p = 100. #[bar]
T = 623.15 #[K]
R = .083145
# Calculation
V = R*T/p # [l/mol] molar volume
v = V/18.0153
# Result
print "Specific volume = ",v," m**3/kg."
```

In [3]:

```
# solution
# Variables
p = 4. #[bar]
T = 773.15 #[K]
R = .083145
V = R*T/p # [l/mol] molar volume
print "Molar volume = ",V," l/mol. "
# Calculation
# using appendix III
# calculating Tc and Pc of different gases according to their mass fractions
Tc1 = .352*32.20 # H2
Tc2 = .148*190.56 # methane
Tc3 = .128*282.34 #ethylene
Tc4 = .339*132.91 # CO
Tc5 = .015*304.10 # CO2
Tc6 = .018*126.09 # N2
Tc = Tc1+Tc2+Tc3+Tc4+Tc5+Tc6 # Tc of gas
# similarly finding Pc
Pc1=.352*12.97
Pc2=.148*45.99
Pc3=.128*50.41
Pc4=.339*34.99
Pc5=.015*73.75
Pc6=.018*33.94
Pc=Pc1+Pc2+Pc3+Pc4+Pc5+Pc6 # Pc of gas
a = (27*R**2*Tc**2)/(64*Pc) # [bar/mol**2]
b = (R*Tc)/(8*Pc) # l/mol
# substituting these values in vanderwall eq and solving by Newton Rhapson method we get
V = 15.74 # [l/mol]
# Result
print "by Vanderwall eq molar volume = ",V," l/mol"
```

In [26]:

```
# solution
# Variables
m = 6.5065 #[g] mixture (basis)
Pv = 2.175 #[kPa] V.P. of water over KOH
Pa = 102.5-2.175 #[kPa] Partial P of n-butane and 1butene
V = 415.1*10**-3 #[l]
R = 8.314472
T = 296.4 #[K]
# Calculation
n = (Pa*V)/R*T # moles of butene and butane
n1 = n*.431 # n-butane
m1 = n1*58 # [g]
n2 = n-n1 # 1 butene
m2 = n2*56 #[g]
m3 = m-m1 # [g] furfural
n3 = m3/96.
# Result
print "component mol percent mass percent"
print " n-Butane ",n1/n*100," ",m1/m*100
print " 1-Butene ",n2/n*100," ",m2/m*100
print " Furfural %.4f"%(n3/n*100)," ",m1/m*100
```

In [27]:

```
# solution
# Variables
P = 5.7+1.01 #[bar] absolute total P
# Calculation
# using Roult's law
vp = 3.293*.7737 #[kPa] vap P of furfural
# using Dalton's law of partial P
n1 = vp/(P*100) # mol fraction of furfural
n2 = 1-n1 # mol fraction of 1 -butene
# Result
print "mol fraction of Furfural is ",n1,"mol fraction of 1-Butene is ",n2,"."
```

In [28]:

```
# solution
# Variables
P = 100. #[kPa] total P
Pw = 2.5326 #[kPa] V.P> of water at dew point
# Calculation
#absolute humity = mass of water vapour/ mass of dry air
H = (Pw/(P-Pw))*(18.0153/28.9697) # absolute humidity
# Result
print "absolute humidity = ",H,"."
```

In [29]:

```
# solution
# Variables
#Ti-Tf = mu*(Pi-Pf)
Pi = 20.7 #[bar]
Pf = 8.7 # [bar]
mu = 1.616 #[K/bar]
Ti = 355.15 #[K]
# Calculation
Tf = Ti-mu*(Pi-Pf)
# Result
print "Outlet temperature is ",Tf," K"
```