# Chapter 4:Solutions Nonelectrolytes¶

## Example 4.1 , Page no:69¶

In :
import math
from __future__ import division

#initialisation of variables
m= 164.2 #gms
M= 60 #gms
V= 0.8 #lit
d= 1.026 #g/cc
mw= 18.02 #gms

#CALCULATIONS
M1= m/M
n= M1/V
G= V*1000*d
G1= G-m
m1= M1*1000/G1
n1= G1/mw
x= M1/(M1+n1)
y= 1-x
p= x*100
p1= y*100
P2= m*100/G

#RESULTS
print"molarity=",round(n,3),"M";
print"molality=",round(m1,3),"m";
print"mole fraction of solute=",round(x,4);
print"mol per cent of solute=",round(p,2),"per cent";
print"mol per cent of solvent=",round(p1,2),"per cent";
print"mol per cent acetic acid by weight=",round(P2,2),"per cent";

molarity= 3.421 M
molality= 4.168 m
mole fraction of solute= 0.0699
mol per cent of solute= 6.99 per cent
mol per cent of solvent= 93.01 per cent
mol per cent acetic acid by weight= 20.0 per cent


## Example 4.2 , Page no:72¶

In :
import math
from __future__ import division

#initialisation of variables
m= 0.0346 #gms
V= 800 #ml
P= 742 #mm
M= 32 #gms
p= 400 #mm

#CALCULATIONS
c= m*1000/V
g= c*760/(P*M)
K= g*22.4
k= c/P
c1= k*p

#RESULTS
print"concentration of oxygen=",round(c,4),"gram per litre";
print"moles dissolved =",round(g,4),"moles";
print"Bunsen absorption =",round(K,4),"litre";
print"grams of oxygen dissolved =",round(c1,4),"gram per litre";

concentration of oxygen= 0.0433 gram per litre
moles dissolved = 0.0014 moles
Bunsen absorption = 0.031 litre
grams of oxygen dissolved = 0.0233 gram per litre


## Example 4.3 , Page no:74¶

In :
import math
from __future__ import division

#initialisation of variables
mn= 0.0134 #gms
mo= 0.0261 #gms
mh= 0.0081 #gms
T= 30 #C
P= 3 #atm
r= 4/5

#CALCULATIONS
V= mn*(273+T)*1000/273
V1= V*r
V2= V1*P
V3= mo*(273+T)*(1-r)*P*1000/273
V4= mh*(273+T)*r*1000/273
V5= V4*P
V6= V2-V1
V7= V5-V4

#RESULTS
print"volume of oxygen=",round(V,1),"ml";
print"volume of nitrogen=",round(V3,1),"ml";
print"volume of helium =",round(V5,1),"ml";
print"volume of nitrogen and helium would be expelled =",round(V7,1),"ml";

volume of oxygen= 14.9 ml
volume of nitrogen= 17.4 ml
volume of helium = 21.6 ml
volume of nitrogen and helium would be expelled = 14.4 ml


## Example 4.4 , Page no:78¶

In :
import math
from __future__ import division

#initialisation of variables
p= 214 #mm
M= 112.5 #gms
m= 18 #gms
m1= 10 #gms

#CALCULATIONS
P= 760-p
M1= m1*P*m/(p*M)

#RESULTS
print"quantity of water=",round(M1,2),"gms";

quantity of water= 4.08 gms


## Example 4.5 , Page no:81¶

In :
import math
from __future__ import division

#initialisation of variables
p = 17.4 #mm
m= 1000 #gms
M= 18 #gms
n= 2 #moles

#CALCULATIONS
P= p*((m/M)/((m/M)+n))
P1= p*(n/((m/M)+n))
dp= p-P1

#RESULTS
print"vapour pressure of solution=",round(P1,2),"mm";

vapour pressure of solution= 0.6 mm


## Example 4.6 , Page no:81¶

In :
import math
from __future__ import division

#initialisation of variables
m= 92.13 #gms
M= 78.11 #gms
n= 1 #moles
p= 119.6 #mm
p1= 36.7 #mm

#CALCULATIONS
n1= m/M
x= n/(n+n1)
y= 1-x
P= y*p
P1= x*p1
P2= P+P1
m1= P/P2
m2= 1-m1

#RESULTS
print"mole fraction of benzene=",round(m1,3);
print"mole fraction of toulene=",round(m2,3);

mole fraction of benzene= 0.794
mole fraction of toulene= 0.206