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My name is Ann [436]
3 years ago
6

A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is th

e enthalpy of vaporization of benzene at 25◦C?
1. 30.8 kJ/mol
2. 13.4 kJ/mol
3. 43.0 kJ/mol
4. 6.09 kJ/mol
5. 25.6 kJ/mol
Chemistry
1 answer:
grandymaker [24]3 years ago
8 0

Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

Number of moles = 95 g/ 78 g/mol

Number of moles = 1.218 mol

Enthalpy of vaporization =  37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

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Answer:- 38.2 g.

Solution:- The equation used for solving this type of calorimetry problems is:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is the change in temperature.

Water temperature is increasing from 14.5 degree C to 50.0 degree C.

\Delta T=50.0-14.5  = 35.5 degree C

q is given as 5680 J and specific heat value is 4.186\frac{J}{g.^0C} .

The equation could be rearranged for m as:

m=\frac{q}{c*\Delta T}

Let's plug in the values in it:

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