The vapor pressure of ethanol is 30°c at 98.5 mmhg and the heat of vaporization is 39.3 kj/mol. determine the normal boiling point of ethanol from this data.

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

c.

explanation:

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer