Question

The molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar enthalpy of fusion of ice is 6.009 kJ/mol. The molar mass of water is 18.02 g/mol. a. How much energy is absorbed when 30.3 g of liquid water boils? b. An energy unit often encountered is the calorie (4.18 J = 1 calorie). Determine the molar enthalpy of fusion of ice in calories per gram.

Answer 1

Answer:

a) $Q = 68.587\,kJ$, b) $L_{f} = 79.778\,\frac{cal}{g}$

Explanation:

a) The energy absorbed by the water is:

$Q = (30.3\,g)\cdot (\frac{1\,mol}{18.02\,g} )\cdot (40.79\,\frac{kJ}{mol} )$

$Q = 68.587\,kJ$

b) The molar enthalpy of fusion of ice is:

$L_{f} = (6.009\,\frac{kJ}{mol} )\cdot (\frac{1000\,J}{1\,kJ} )\cdot (\frac{1\,mol}{18.02\,g} )\cdot (\frac{1\,cal}{4.18\,J} )$

$L_{f} = 79.778\,\frac{cal}{g}$

Answer 2

The energy is absorbed  and the molar enthalpy of fusion of ice  is mathematically given as

Q = 68.587kJ

Lf=79.778cal/g

What is energy is absorbed when 30.3 g of liquid water boils and the molar enthalpy of fusion of ice in calories per gram.?

Question Parameter(s):

The molar enthalpy of vaporization of water is 40.79 kJ/mol,

the molar enthalpy of fusion of ice is 6.009 kJ/mol.

The molar mass of water is 18.02 g/mol.

a)

Generally, The equation energy absorbed by the water  is mathematically given as

Q=mw*1/molar mass* molar enthalpy of vaporization of water

Q = (30.3g)* 1/18.0  *40.79

Q = 68.587kJ

b)

the molar enthalpy of fusion of ice in calories per gram.

Lf = 6.009*1000*(1/18.02)*(1/4.18)

Lf=79.778cal/g

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