Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :
The expression used will be:
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene =
= specific heat of liquid benzene =
= enthalpy change for vaporization =
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
Therefore, the energy removed must be, -67.7 kJ
1.04 ⨯ 10 M
<em>A</em> = <em>ε</em> by the Beer-Lambert law, where
<em>A</em> and <em>ε </em>are the same for both solutions. Therefore, is constant;
is inversely proportional to
. The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.
The 13.0 mL standard solution has a concentration of 5.17 ⨯ M. Diluting it to 45.0 mL results in a concentration of
1.494 M.
is inversely related to
for the two solutions. As a result, c₂ =
3.126 M.
The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of 1.04 ⨯
M.