Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
The number of C2H5OH in a 3 m solution that contain 4.00kg H2O is calculate as below
M = moles of the solute/Kg of water
that is 3M = moles of solute/ 4 Kg
multiply both side by 4
moles of the solute is therefore = 12 moles
by use of Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 12 moles
=12 moles/1 moles x 6.02 x10^23 = 7.224 x10^24 molecules
I would say developed technologies which somehow were lost
like the stone age: the stone hammer, who used that afterwards in the roman empire era and stuff?
