Respuesta :
Explanation:
First, we will calculate the number of moles of benzene as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{125 g}{78.11 g}[/tex]
= 1.60 moles
We assume that the heat energy is removed from the benzene in three separate steps:
1) In cooling the gaseous sample from 425K to 353K (its boiling point)
2) In liquefying the sample at 353 K
3) In cooling the liquid from 353 K to 335 K
So, we will first calculate the energy removed from the benzene at each step, then add them together to find the total amount of energy removed from the benzene.
1) Q = [tex]m \times C \times \Delta T[/tex]
= 125 x 1.06 x (425 - 353)
= 9540J
or, = 9.540 kJ (as 1 kJ = 1000 J)
2) Also, Q = [tex]\Delta H_{vap} \times 1.6 mol[/tex]
= 33.9 x 1.6
= 54.24 kJ
3) Q = [tex]m \times C \times \Delta T[/tex]
= 125 x 1.73 x (353 - 335)
= 3893J
= 3.893kJ
Hence, total = 9.540 + 54.240 + 3.893 = 67.673 kJ
Thus, we can conclude that 67.673 kJ energy must be removed from a 125 g given sample of benzene under the given conditions.
67.673 kJ energy must be removed from a 125 g given sample of benzene under the given conditions.
Number of moles:
It is defined as the ratio of given mass over molar mass.
Given:
Mass = 125g
Molar mass =78.11g/mol
[tex]\text{Number of moles}=\frac{\text{given mass}}{\text{molar mass}}\\\\ \text{Number of moles}=\frac{125g}{78.11g/mol}\\\\ \text{Number of moles}=1.60 moles[/tex]
Calculation of heat energy:
The heat energy is removed from the benzene in three separate steps:
- In cooling the gaseous sample from 425K to 353K (its boiling point).
- In liquefying the sample at 353 K.
- In cooling the liquid from 353 K to 335 K.
1) In order to calculate the energy we will use specific heat formula which is given as:
[tex]Q=m*c*\triangle T[/tex]
Substituting values in this formula
[tex]Q=m*c*\triangle T\\\\ Q= 125 * 1.06 * (425 - 353)\\\\ Q= 9540J=9.540kJ[/tex]
2)
[tex]Q = H_{vap}*1.6mol\\\\Q=33.9*1.6\\\\Q=54.24kJ[/tex]
3)
[tex]Q=m*c*\triangle T\\\\Q=125*7.73*(353-335)\\\\Q=3893J=3.893kJ[/tex]
Thus, total [tex]Q= 9.540 + 54.240 + 3.893 = 67.673 kJ[/tex]
We can conclude that 67.673 kJ energy must be removed from a 125 g given sample of benzene under the given conditions.
Find more information about Heat energy here:
brainly.com/question/13439286
